Introduction

The term magnetism comes from Magnesia, the name of an ancient city in Asia Minor, where the Greeks found certain very unusual stones more than 2000 years ago. These stones, called lodestones, possess the unusual property of attracting pieces of iron. Such magnets were first fashioned into compasses and used navigation by the Chinese in the twelfth century A.D.

In the sixteenth century, William Gilbert, Queen Elizabeth’s physician, made artificial magnets by rubbing pieces of iron against lodestones. He suggested that a compass always points north and south because the earth itself shows the magnetic properties. Later in 1750, John Michaell in England found that magnetic poles obey the inverse-square law, and his results were confirmed by Charles Coulomb. The subjects of magnetism and electricity developed almost independently until 1:20, when a Danish physicist named Hans Christian Oersted discovered, in a classroom demonstration, that an electric current affects a magnetic compass. He saw that magnetism was related to electricity. Shortly thereafter, the French physicist Andre Marie Ampere proposed that electric currents are the source of all magnetic phenomena. This chapter is all about how electricity is connected with magnetism.

PERMANENT MAGNETS

Magnets produced from magnetic materials are called Artificial magnets. They can be made in a variety of shapes and sizes and are used extensively in electrical apparatus.

Artificial magnets are generally made from special iron or steel alloys which are usually magnetized electrically. The material to be magnetized is inserted into a coil of insulated wire and a heavy flow of electrons is passed through the wire. Magnets can also be produced by stroking a magnetic material with magnetite or with another artificial magnet. The forces causing magnetization are represented by magnetic lines of force.

Artificial magnets are usually classified as permanent or temporary, depending on their ability to retain their magnetic properties after the magnetizing force has been removed. Magnets made from substances, such as hardened steel and certain alloys which retain a great deal of their magnetism, are called permanent magnets. These materials are relatively difficult to magnetize because of the opposition offered to the magnetic lines of force as the lines of force try to distribute themselves throughout the material. The opposition that a material offers to the magnetic lines of force is called reluctance. All permanent magnets are produced from materials having a high reluctance.

A material with a low reluctance, such as soft iron or annealed silicon steel, is relatively easy to magnetize but will retain only a small part of its magnetism once the magnetizing force is removed. Materials of this type that easily lose most of their magnetic strength are called temporary magnets. The amount of magnetism which remains in a temporary magnet is referred to as its residual magnetism. The ability of a material to retain an amount of residual magnetism is called the retentivity of the material Magnets are also described in terms of the permeability of their materials, or the ease with which magnetic lines of force distribute themselves throughout the material. A permanent magnet, which is produced from a material with a high reluctance, has a low permeability. A temporary magnet, produced from a material with a low reluctance, would have a high permeability.

PROPERTIES OF BAR MAGNETS

(1) Attractive Property and Poles

When a magnet is dipped into iron fillings it is found that the concentration of iron filings, i.e., attracting power of the magnet is maximum at two points near the ends and minimum at the centre. The places in a magnet where its attracting power is maximum are called poles while the place of minimum attracting power is called the neutral region.

Fig. 6.2

(2) Directive Property and N-S Poles

When magnet is suspended its length becomes parallel to N-S direction. The pole pointing north is called the north pole while the other pointing south is called the south pole.

(3) Opposite poles ($N$ and $S$) attract and like poles ($N$ and $N$, or $S$ and $S$) repel.

Fig. 6.3

(4) Magnetic Axis and Magnetic Meridian

The line joining the two poles of a magnet is called magnetic axis and the vertical plane passing through the axis of a freely suspended or pivoted magnet is called magnetic meridian.

(5) Magnetic Length ($2l$)

The distance between two poles along the axis of a magnet is called its effective or magnetic length. As poles are not exactly at the ends, the effective length is lesser than the actual length of the magnet.

Fig. 6.4

(6) Poles Exist in Pairs

In a magnet the two poles are found to be equal in strength and opposite in nature. If a magnet is broken into number of pieces, each piece becomes a magnet with two equal and opposite poles. This shows that monopoles do not exist.

Fig. 6.5

(7) Repulsion is a Sure Test of Polarity

A pole of a magnet attracts the opposite pole while repels similar pole. A sure test of polarity is repulsion and not attraction, as attraction can take place between opposite poles or a pole and a piece of unmagnetised magnetic material due to ‘induction effect’.

(8) Magnetic Induction

A magnet attracts certain other substances through the phenomenon of magnetic induction i.e., by inducing opposite pole in a magnetic material on the side facing it as shown in fig.

Fig. 6.6

IDEA BOX

Interesting Fact

Seeds of two tomatoes varieties Rocco and Monza were treated by passing them through an artificial magnetic field (MF) with a constant defined velocity before seeding. The seedlings obtained from MF treated and non MF-treated seeds were planted into the MF treated and non MF-treated plots. They were irrigated by MF-treated and non MF treated water.

Observations were made on early-yield, total-yield, beginning of blooming, and quality of fruit. While significant differences were not observed in Rocco, important MF effects were clearly seen on Monza. Yield increases on Monza in magnet treated plots were around $28-51$ percent, especially in early yields, and Monza bloomed three-four days earlier.

Pole strength and magnetic moment of a magnet

The strength of attracting a magnetic material by a magnetic pole is measured by pole strength, generally denoted by $m$.

The north and south poles of a magnet are said to be associated with the pole strengths $+m$ and $-m$, respectively, each of which is a scalar, having the dimensional formula $[A L]$ and SI unit Am. Each pole strength is imagined to be distributed over the entire cross-section of the magnet. Therefore, if the magnet is cut along it’s length into the two equal parts, each part will have it’s poles of the strengths $+m / 2$ and $-m / 2$, respectively, and if the magnet is cut perpendicular to it’s length into the two equal parts, each part will have it’s poles of same initial strengths $+m$ and $-m$, respectively.

Fig. 6.7

The magnetic moment of a magnet is defined as a vector $\vec{M}$ whose magnitude is

$ M=n(2 l) $

where $m$ is the pole strength of magnet and $2 l$ is the effective length of magnet and whose direction is from the south pole to the north pole. It has the dimensional formula $[A L^{2}]$ and SI unit $Am^{2}$. If the magnet is cut along it’s length, or perpendicular to its length, into the two equal parts, each part will have it’s magnetic moment of magnitude one-half of that of the original magnetic moment, as shown above.

MAGNETIC FIELD

The space around a magnet (or a current carrying conductor) in which its magnetic effect can be experienced is called the magnetic field. It is a quantity that has both direction and magnitude. Pictorically it is represented by magnetic lines of forces. Tangent to magnetic lines of forces gives direction of magnetic field at that point. No two magnetic lines of force intersect each other.

The magnetic field in a region is said to be uniform if the magnitude of its strength and direction is same at all points in that region.

Fig. 6.8

A magnetic field in a region is said to be uniform if the magnitude of its strength and direction is same at all the points in that region. The strength of magnetic field is also known as magnetic induction or magnetic flux density,

The SI unit of strength of magnetic field is Tesla ( $T$ )

1 Tesla $=1$ newton ampere ${ }^{-1}$ metre $^{-1}(NA^{-1} m^{-1})=1$ Weber metre ${ }^{-2}(Wb m^{-2})$

The cgs unit is gauss $(G), 1$ gauss $(G)=10^{-4}$ Tesla $(T)$.

COULOMB’S LAW IN MAGNETISM

This law defines the force $F$ between the two magnetic poles of the strengths $m_1, m_2^{\prime}$ at a distance $r$ apart in a medium as

$ F=(\frac{\mu}{4 \pi})(\frac{m_1 m_2}{r^{2}}) (SI) $

$ F=(\frac{1}{\mu})(\frac{m_1 m_2}{r^{2}}) (CGS) $

where $\mu$ is the permeability of medium and $\mu=\mu_0=4 \pi \times 10^{-7} N / A^{2}$ in SI and $\mu=\mu_0=1$ (unitless) in CGS. Evidently, the force $F$ is directly proportional to the product of the pole strengths and inversely proportional to the square of the distance $r$. It acts along the line joining the two magnetic poles. It is attractive for the unlike magnetic poles and repulsive for the like magnetic poles. It depends on the medium in which the magnetic poles are situated.

CHECK POINT

A friend tells you that a refrigerator door, beneath its layer of white-painted plastic, is made of aluminum. How could you check to see if this is true (without any scraping) ?

Show Answer

SOLUTION:

I will bring a magnet and make it touch with the refrigerator door. If the magnet shows a feeble attraction with it then I can say that it is aluminum.

ILLUSTRATION-6.1

Four identical bar magnets, each of magnetic moment $M$, are arranged to form a square, as shown. Calculate the resultant magnetic moment of the system.

Show Answer

SOLUTION:

The magnetic moments of the given four identical bar magnets are acting along the sides of a square, as shown. The resultant of each of the two pairs of magnetic moments, i.e., $(1,4)$ and $(2,3)$, is $M \sqrt{2}$ and it is represented by the same diagonal of the square. Therefore, the resultant of all the four magnetic moments, taken together, is $2 \sqrt{2} M$, which is acting along the diagonal.

Fig. 6.10

MAGNETIC FIELD DUE TO MAGNETIC POLE

In SI, the magnetic induction $\overrightarrow{B}$ due to a magnetic pole is defined at a point at a distance $r$ as the force per unit north pole placed there, having the magnitude

$\quad$ $B=\left(\frac{\mu}{4 \pi}\right)\left(\frac{m}{r^2}\right) \quad$ tesla

$\Rightarrow \quad H=\frac{B}{\mu}=\left(\frac{1}{4 \pi}\right)\left(\frac{m}{r^2}\right) \quad \mathrm{A} / \mathrm{m}$

and direction away for north pole and towards for the south pole. In CGS, the magnetic field intensity $\vec{H}$ due to a magnetic pole is defined at a point at a distance $r$ as the force per unit north pole placed there, having the magnitude

$ H=(\frac{1}{\mu})(\frac{m}{r^{2}}) \text{ oersted } \Rightarrow B=\mu H=(\frac{m}{r^{2}}) \text{ gauss } $

and direction away for the north pole and towards for the south pole.

FORCE ON MAGNETIC POLE PLACED IN MAGNETIC FIELD

In SI, the force a magnetic pole of strength $m$ placed in a magnetic field $\vec{B}$ is

$ \vec{F}= \pm m \vec{B} $

where + sign is for the north pole and - sign is for the south pole. Thus, the force is in the direction of $\vec{B}$ for the north pole and in the opposite direction for the south pole (Fig. 6.11).

In CGS, the force on a magnetic pole of strength $m$ placed in a magnetic field $\vec{H}$ is

$ \vec{F}= \pm m \vec{H} $

where + sign is for the north pole and - sign is for the south pole. Thus, the force is in the direction of $\vec{H}$ for the north pole and in the opposite direction for the south pole. (Fig. 6.11)

MAGNETIC FIELD DUE TO A BAR MAGNET

End-on Position

The magnetic field $\vec{B}$ at a point $P$ on the axis of a bar magnet at distance $r$ form it’s mid point $O$ is given by

$\begin{aligned} \vec{B} & =\vec{B}_N+\vec{B}_S \\ \Rightarrow \quad B & =B_N-B_S\end{aligned}$

$=\left(\frac{\mu}{4 \pi}\right) \frac{m}{(r-l)^2}-\frac{\mu}{4 \pi} \frac{m}{(r+l)^2}$

$=\left(\frac{\mu}{4 \pi}\right) \frac{2 M r}{\left(r^2-l^2\right)^2} \approx\left(\frac{\mu}{4 \pi}\right)\left(\frac{2 M}{r^3}\right)$ for $l \ll r$

which is having the direction parallel to the magnetic moment $\vec{M}$ of the bar magnet.

Fig. 6.13

Broadside-on Position

The magnetic field $\vec{B}$ at a point $P$ on the equator of the bar magnet at a distance $r$ from it’s mid point $O$ is given by

$\begin{aligned} \vec{B} & =\vec{B}_N+\vec{B}_S \\ \Rightarrow \quad B & =2 B_N \cos \theta\end{aligned}$

$\quad$ $\quad$ $=2 B_{\mathrm{s}} \cos \theta$

$=2\left(\frac{\mu}{4 \pi}\right) \frac{m}{\left(r^2+l^2\right)} \cdot \frac{l}{\sqrt{\left(r^2+l^2\right)}}$

$=\left(\frac{\mu}{4 \pi}\right) \frac{M}{\left(r^2+l^2\right)^{3 / 2}} \approx\left(\frac{\mu}{4 \pi}\right)\left(\frac{M}{r^3}\right)$ for $l \ll r$

which is having the direction antiparallel to the magnetic moment $\vec{M}$ of the bar magnet. Evidently, the magnetic field at the broadside-on position is one-half of that at the end-on position for the same distance $r$ from the broadside-on position is one-half of that at the end-on position for the same distance $r$ from the mid point $O$ of the bar magnet.

Fig. 6.14

Any Other Position

The magnetic moment $\vec{M}$ of bar magnet has the two components, namely, $M \cos \theta$ and $M \sin \theta$, as shown. The point $P$ is neither on the axis nor on the equator of bar magnet but having any other position at a distance $r$ from the mid point $O$. It’s position is end-on for the component $M \cos \theta$ and broadside-on position for the component $M \sin \theta$. Therefore, the magnetic field $\vec{B}$ at the point $P$ is given by

$ \begin{aligned} \vec{B} & = \vec{B} _r+ \vec{B} _{\theta} \\ \Rightarrow \quad B & =\sqrt{ }(B_r{ }^{2}+B _{\theta}{ }^{2}) \\ & =\sqrt{ } \biggl[\biggl(\frac{\mu}{4 \pi}\biggl)^{2}\biggl(\frac{2 M \cos \theta}{r^{3}}\biggl)^{2}+\biggl(\frac{\mu}{4 \pi}\biggl)^{2}\biggl(\frac{M \sin \theta}{r^{3}}\biggl)^{2}\biggl] \\ & =\biggl(\frac{\mu}{4 \pi}\biggl)\biggl(\frac{M}{r^{3}}\biggl) \sqrt{(4 \cos ^{2} \theta+\sin ^{2} \theta)} \\ & =\biggl(\frac{\mu}{4 \pi}\biggl) \biggl(\frac{M}{r^{3}}\biggl) \sqrt{(1+3 \cos ^{2} \theta)} \end{aligned} $

whose direction is inclined at an angle $(\phi+\theta)$ with the magnetic moment $\vec{M}$, where

$ \tan \phi=\frac{B _{\theta}}{B_r}=(\frac{1}{2}) \tan \theta $

Fig. 6.15

FORCE AND TORQUE ON BAR MAGNET PLACED IN MAGNETIC FIELD

Uniform Magnetic Field

When a bar magnet is placed in a uniform magnetic field $\vec{B}$ at an angle $\theta$, the total force one it becomes

$\vec{F}= \vec{F} _N+\bar {F}_S=(+m)(+m) \vec{B}+(-m) \vec{B} = 0 $

and hence, it does not undergo any translatory motion.

The total torque on the bar magnet due to the magnetic field is

$ \vec{\tau}=\vec{M} \times \vec{B} $

whose magnitude is

$ \begin{aligned} \tau & =(m B) l \sin \theta+(m B) l \sin \theta \\ & =(2 m l) B \sin \theta \end{aligned} $

$ =M B \sin \theta $

and whose direction is perpendicular to the plane of figure and inwards, as shown. This torque produces the rotatory motion of the bar magnet in the clockwise direction.

Non-uniform Magnetic Field

When the bar magnet is placed in a non-uniform magnetic field $\vec{B}$ at an angle $\theta$, the total force on it becomes

$ \begin{aligned} \vec{F} & = \vec{F} _N+ \vec{F} _S \\ & =(+m) \vec{B}+(-m) \overrightarrow{B^{\prime}}=m(\vec{B}- \vec{B} ^{\prime}) \neq 0 \end{aligned} $

which produces the translatory motion of the bar magnet in its’ direction.

The total torque on the bar magnet due to the magnetic field is

$ \vec{\tau}=\vec{M} \times \vec{B} _{a v} $

whose magnitude is

$ \begin{aligned} \tau & =(m B) l \sin \theta+(m B^{\prime}) l \sin \theta \\ & =(2 m l)\bigg(\frac{B+B^{\prime}}{2}\bigg) \sin \theta=M B _{a v} \sin \theta \end{aligned} $

where $B _{a v}$ is the average magnetic field, and whose direction is perpendicular to the plane of figure and inwards, as shown. This torque produces the rotatory motion of the bar magnet in the clockwise direction.

POTENTIAL ENERGY OF BAR MAGNET PLACED IN MAGNETIC FIELD

The potential energy of a bar magnet of the magnetic moment $\vec{M}$ placed in a magnetic field $\vec{B}$ is

$ \begin{aligned} U & =U_N+U_S \\ & =(+m) V_N+(-m) V_S \\ & =(-m)(V_S-V_N) \\ & =(-m)(B \cos \theta) 2 l \\ & =-M B \cos \theta \\ & =-\vec{M} \cdot \vec{B} \end{aligned} $

Fig. 6.18

where $U_N, U_S$ are the potential energies of the north pole and south pole of the bar magnet, respectively, and $V_N, V_S$ are the magnetic potentials at the positions of the two poles, respectively. Evidently, $U$ may be positive, or zero, or nagative, depending on the angle $\theta$. $U$ is maximum and equal to $+M B$ for $\theta=180^{\circ}$ and minimum and equal to $-M B$ for $\theta=0^{\circ}$. Therefore, the position $\theta=0$ of the bar magnet is that of the stable equilibrium.

WORK DONE IN ROTATING BAR MAGNET IN MAGNETIC FIELD

The work done in rotating a bar magnet of the magnetic moment $\vec{M}$ from it’s initial position $\theta_I$ to a final position $\theta_F$ in a magnetic field $\vec{B}$ is

$ W=U_F-U_I=(-M B \cos \theta_F)-(-M B \cos \theta_I)=M B(\cos \theta_I-\cos \theta_F) $

Fig. 6.19

For example, when the bar magnet is rotated through $180^{\circ}$ from the position of stable equilibrium $(\theta=0^{\circ})$, then the work done is

$W=M B(\cos 0^{\circ}-\cos 180^{\circ})=2 M B$

CHECK POINT:

One way to make a compass is to stick a magnetized needle into a piece of cork and to float it in a glass bowl full of water. The needle will align itself with the horizontal component of the earth’s magnetic field. Since the north pole of this compass is attracted northward, will the needle float toward the north side of the bowl ? Defend your answer.

Show Answer

SOLUTION:

Yes, the north pole of the needle will float toward the north side of the bowl. The earth is a huge magnet whose north pole is towards the geographic south and the south pole is towards the geographic north. So, the north pole of the needle will align towards the south pole of the earth’s magnet which is northward. Therefore, the north pole of the needle will tend to float toward the north side of the bowl.

ILLUSTRATION-6.2

The magnetic field due to a small bar magnet at a distance of $24 cm$ on its equator is $B$ and the magnetic field at a distance $d$ on it’s axis is $16 B$. Then, what is the distance $d$ ?

SOLUTION:

The magnetic field due to the small bar magnet of the magnetic moment $M$ at a distance of $0.24 m$ on its equator is

$ B=(\frac{\mu_0}{4 \pi}) \frac{M}{(0.24)^{3}} $

The magnetic field at a distance $d$ on the axis of magnet is $16 B=(\frac{\mu_0}{4 \pi}) \frac{2 M}{(d)^{3}}$

Dividing the two equations, we have

$\begin{aligned} 16 & =2\left(\frac{0.24}{d}\right)^3 \\ \Rightarrow \quad d & =0.12 \mathrm{~m}=12 \mathrm{~cm} .\end{aligned}$

Terresterial Magnetism

William Gilbert suggested that earth itself behaves like a huge magnet. This magnet is so oriented that its $S$ pole is towards geographic north and $N$ pole is towards the geographic south.

The earth behaves as a magnetic dipole inclined at small angle $11.5^{\circ}$ to the earth’s axis of rotation with its south pole pointing geographic north. The idea of earth having magnetism is supported by following facts.

(a) A freely suspended magnet always comes to rest in $N$ - $S$ direction. (b) A piece of soft iron buried in $N-S$ direction inside the earth acquires magnetism. (c) Existence of neutral points. When we draw field lines of bar magnet we get neutral points where magnetic field due to magnet is neutralized by earth’s magnetic field.

The magnetic field at the surface of earth ranges from nearly $30 \mu T$ near equator to about $60 \mu T$ near the poles. The magnetic field on the axis is nearly twice the magnetic field on the equatorial line.

Cause of earth’s magnetism

Sir William Gilbert first suggested the existence of a powerful magnet inside the earth. This is not possible because.

(a) Temperature inside earth is so high that it will not be possible for magnet to retain magnetism.

(b) If there was a magnet inside the earth then position of earth’s magnetic poles would have not changed.

(c) The process of magnetisation of this magnet is not understood.

Grover suggested that earth’s magnetism is due to flow of current near outer surface of earth. These currents are produced due to sun. The hot air rising from regions near equator while going towards north and south hemispheres gets electrifield. These then magnetise ferromagnetic materials near the surface of earth.

According to another view earth’s core has many conducting materials like iron and nickel in molten state. Conventional currents are produced in this semi fluid core due to rotation of earth about its axis which generates magnetism.

Another view says magnetism is due to presence of ionised gases in atmosphere. The high energy sun rays ionize gas atoms in upper layer of atmosphere. The radioactivity of atmosphere and cosmic rays also ionize the gases. Strong electric currents flow due to rotation of earth producing magnetism.

Thus most likely cause of earth’s magnetism is the motion and distribution of charged materials in and outside the earth.

Some definitions

Geographic Axis

It is straight line passing through the geographic poles of the earth. It is the axis of rotation of the earth. It is known as polar axis.

Geographic Meridian

It is a vertical plane passing through geographic north and south poles of the earth.

Geographic Equator

A great circle on the surface of the earth in a plane perpendicular to geographical axis is called geographic equator. All places on geographic equator are at equal distances from geographical poles.

Magnetic Axis

It is a straight line passing through the magnetic poles of the earth. It is inclined to geographic axis at nearly $11.5^{\circ}$.

Fig. 6.20

Magnetic Meridian

It is a vertical plane passing through the magnetic north and south poles of the earth.

Magnetic Equator

A great circle on the surface of the earth in a plane perpendicular to magnetic axis is called magnetic equator. All places on magnetic equator are at equal distance from magnetic poles.

COMPONENTS OF EARTH’S MAGNETIC FIELD

The resultant magnetic field $\vec{B}$ of earth at a place has the two components, namely, the vertical component $B_V$ and horizontal component $B_H$. The component $B_V$ has the vertical direction which is downwards in the northern hemisphere and upwards in the southern hemisphere. The component $B_H$ has the horizontal direction which is from the south to the north everywhere.

$ \begin{aligned} B_V & =B \sin \theta \\ B_H & =B \cos \theta \\ B & =\sqrt{(B_V^{2}+B_H^{2})} \\ \tan \theta & =\frac{B_V}{B_H} \end{aligned} $

The vertical component $B_V=0$ at the magnetic equator and the horizontal component $B_H=0$ at the magnetic north and south poles of earth. At Delhi, $B_H=35 \mu T=0.35 G$.

Fig. 6.21

Angle of Dip, or Inclination, $\theta$

At a place, the angle $\theta$ which the resultant magnetic field $\vec{B}$ of earth makes with the horizontal is called the angle of dip, or inclination. A freely-suspended magnet will keep it’s axis at this angle $\theta$ with the horizontal, the south pole being above the north pole in the northern hemisphere and the south pole being below the north pole in the southern hemisphere. It is $0^{\prime}$ at the magnetic equator and $90^{\circ}$ at the magnetic north and south poles of earth. At other places, it lies between $0^{\circ}$ and $90^{\circ}$. At Delhi, it is $42^{\circ} 12.7^{\circ}$. The angle of dip $\theta$ measured in the magnetic meridian is called the true angle of dip and the angle of dip $\theta^{\prime}$ measured in any other vertical plane inclined at an angle $\phi$ with the magnetic meridian is called the apparent angle of dip. The ‘Dip Circle’ is used to measure the angle of dip in any vertical plane.

$\begin{aligned} \tan \theta & =\frac{B_V}{B_H} \\ \tan \theta^{\prime} & =\frac{B_V}{B_H^{\prime}}=\frac{B_V}{B_H \cos \phi}\end{aligned}$

$\begin{aligned} & \Rightarrow \quad \tan \theta^{\prime}=\left(\frac{1}{\cos \phi}\right) \tan \theta \\ & \Rightarrow \quad \theta^{\prime}>\theta\end{aligned}$

Variation, or Declination, $\alpha$

At a place, the angle $\alpha$ between the magnetic and geographic meridians is called the variation, or declination. It is $0^{\circ}$ at Delhi.

Magnetic Elements of Earth

The three parameters, namely, the angle of dip or inclination $\theta$, variation or

declination $\alpha$ and horizontal component $B_H$ of earth’s magnetic field $\vec{B}$, are together called the magnetic elements of earth. These elements completely define the earth’s magnetism.

Magnetic Compass

Present day magnetic compasses use the same forces that guided ancient mariners. A magnetized needle, in conjunction with a compass card, rotates horizontally. Present day compasses are superior to the ancient ones through a heightened knowledge of magnetic laws and greater precision in construction.

The Earth’s magnetic lines of force provide the directional information needed to navigate. A compass detects and converts the energy from these magnetic lines of force into a directional display. In order to understand the operation of a ship’s compass, it is first necessary to understand some basic information about the Earth’s magnetic field.

Fig. 6.23

NEUTRAL POINTS

Points where net magnetic field is zero are called neutral points.

Magnet in Horizontal Position

When a magnet is placed in the horizontal position with it’s north pole facing the north, we get the two neutral points $P_1$ and $P_2$on the equator of magnet, symmetrically placed with respect to the magnet, as shown. At the neutral points, the earth’s horizontal component $ \vec{B} _H$ and the magnetic field $\vec{B}$ of magnet balance each other so that the resultant magnetic field is zero there. Therefore,

$ \begin{aligned} B_H=B & =\biggl(\frac{\mu}{4 \pi}\biggl) \frac{M}{(r^{2}+l^{2})^{3 / 2}} \\ & \approx\biggl(\frac{\mu}{4 \pi}\biggl) \frac{M}{r^{3}} \text{ for } l \ll r \end{aligned} $

When the magnet is placed in the horizontal position with it’s north pole facing the south, we get the two neutral points $P_1$ and $P_2$ on the axis of magnet, symmetrically placed with respect to the magnet, as shown. At the neutral points, we have

$ \begin{aligned} B_H & =B=\biggl(\frac{\mu}{4 \pi}\biggl) \frac{2 M r}{(r^{2}-l^{2})^{2}} \\ & \approx\biggl(\frac{\mu}{4 \pi}\biggl) \frac{2 M}{r^{3}} \text{ for } l \ll r \end{aligned} $

Magnet in Vertical Position

When the magnet is placed in the vertical position with it’s north pole resting on the table, we get only one neutral point $P$ on the south of magnet in the plane of the table top, as shown. At the neutral point we have

$ \begin{aligned} B_H & =B=B_N-B_S \cos \theta & & \text{ (for short magnet) } \\ & \approx B_N & & \text{ (for long magnet, i.e., } l \gg r) \end{aligned} $

where,

$ \begin{aligned} & B_N=\biggl(\frac{\mu}{4 \pi}\biggl) \frac{m}{r^{2}} \\ & B_S=\biggl(\frac{\mu}{4 \pi}\biggl) \frac{m}{(r^{2}+4 l^{2})} \end{aligned} $

Fig. 6.27

are the magnetic fields due to the two poles, taken separately.

When the magnet is placed in the vertical position with it’s south pole resting on the table, we get only one neutral point $P$ on the north of magnet in the plane of the table top, as shown. At the neutral point, we have

$ B_H=B=B_S-B_N \cos \theta \text{ (for short magnet) } $

$ \cong B_S(\text{ for long magnet, i.e., } l>r) $

where $\quad B_N=(\frac{\mu}{4 \pi}) \frac{m}{(r^{2}+4 l^{2})}$

$ B_s=(\frac{\mu}{4 \pi}) \frac{m}{r^{2}} $

are the magnetic fields due to the two poles.

Fig. 6.28

CHECK POINT:

Magnet $A$ has twice the magnetic field strength of magnet $B$, and, at a certain distance, it pulls on magnet $B$ with a force of $50 N$. With how much force, then, does magnet $B$ pull on magnet $A$ ?

Show Answer

SOLUTION:

The magnet $B$ will pull on magnet $A$ with the same force as magnet $A$ pulls on magnet $B$, i.e, with a force of $50 N$ only because the force of attraction between the two magnets is a combined property of both the magnets and acts equally on both.

ILLUSTRATION-6.3

A bar magnet of the length of $0.2 m$ is placed horizontally with it’s $S$-pole towards the geographic north. The two neutral points are situated on the axis of magnet at the distance of $0.4 m$ from the centre of magnet. If the horizontal component of earth’s magnetic field is $B_H=3.2 \times 10^{-5} T$, then find the pole strength of magnet.

Show Answer

SOLUTION:

The two neutral points are situated in the end-on positions on the axis of magnet such that the magnetic field $B$ of magnet and earth’s horizontal component $B_H$ are equal but opposite and hence, cancel out each other exactly. Thus

$B = B_H$

$ \begin{aligned} & \Rightarrow \quad\biggl(\frac{\mu_0}{4 \pi}\biggl) \frac{2 M r}{(r^{2}-l^{2})^{2}}=B_H \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad(10^{-7}) \frac{2\times 2ml\times r}{(r^{2}-l^{2})^{2}}=B_H \end{aligned} $

$\Rightarrow \quad(10^{-7}) \frac{2\times 2\times m\times 0.1\times 0.4}{[0.4^{2}-(0.1)^{2}]^{2}}=3.2\times 10^{-5} $

$ \begin{aligned} & m=45 Am \end{aligned} $

MAGNETIC MATERIALS

Certain materials, when placed in a magnetising field $ \vec{B} _0$, i.e., in a magnetising field intensity $\vec{H}= \vec{B} _0 / \mu_0$, get magnetised. Such materials are called the magnetic materials. Some materials get magnetised very strongly, e.g., ferromagnetics and others get magnetised very feebly, e.g., diamagnetics and paramagnetics.

TERMS RELATED TO MAGNETISM

Magnetic intensity $(\vec{H})$

When a magnetic material is placed in a magnetic field, it becomes magnetised. The capability of the magnetic field to magnetise a material is expressed by means of a magnetic vector $\overrightarrow{H}$, called the ‘magnetic intensity’ of the field. The relation between magnetic induction $B$ and magnetising field $\vec{H}$ is $\vec{B}=\mu \vec{H}, \mu$ being permeability of medium.

Intensity of magnetisation (I)

When a material is placed in a magnetising field, it acquires magnetic moment M. The intensity of magnetisation is defined as the magnetic moment per unit volume i.e., $I_m=\frac{M}{V}, V$ being volume of mateiral. If the material is in the form of a bar of cross-sectional area $A$, length 21 and pole strength $m$, then

$M=m \times 2 \ell ; V=A \times 2 \ell \quad \therefore I_m=\frac{M}{V}=\frac{m \cdot 2 \ell}{A \cdot 2 \ell}=\frac{m}{A}$

Magnetic Susceptibility $(\chi)$

The magnetic susceptibility is defined as the intensity of magnetisation per unit magentising field i.e.,

$ \chi=\frac{I_m}{H} $

Magnetic permeability $(\mu)$

The magnetic permeability of a material is the measure of degree to which the material can be permeated by a magnetic field and is defined as the ratio of magnetic induction (B) in the material to the magnetising field i.e.

$\mu=\frac{B}{H}$

RELATION BETWEEN MAGNETIC SUSCEPTIBILITY AND PERMEABILITY

We have magnetic induction in mateiral, B = $\mu H$

Also $B = B_0 + B_m$

where $B_0$ = magnetic induction in vacuum produced by magnetising field

$B_m$ = magnetic induction due to magnetisation of material.

But $B_0=\mu_0 H$ and $B_m=\mu_0 I_m \Rightarrow B=\mu_0\left[H+I_m\right]$

$\therefore \quad B=\mu_0 H\left[1+\frac{I_m}{H}\right]=B_0[1+\chi]; \therefore B/B_0=[1+\chi]$

$\therefore \quad B/B_0=\frac{\mu{H}}{\mu_0{H}}={\mu}/{\mu_0}$, the relative magnetic permeability

$\therefore \mu_r=1+\chi$

This is required relation.

CLASSIFICATION OF MATERIAIS

According to the behaviour of substances in magnetic field, they are classified into three categories

(I) DIAMAGNETIC SUBSTANCES

These are substances which when placed in a strong magnetic field acquire a feeble magnetism opposite to the direction of magnetising field.

The examples are copper gold, antimony, bismuth, alcohol, water, quartz, hydrogen, etc.

Fig. 6.29 : Behaviour of Diamagnetic substance in an external Magnetic Field $ \vec{B} _0$

The characteristics of diamagnetic substances are

(a) They are feebly repelled by a strong magnet

(b) Their susceptibility is negative (i.e. $\chi<0$ )

(c) Their relative permeability is less than 1. (i.e. $\mu_r<1$ )

(d) Their susceptibility is independent of magnetising field and temperature (except for Bismuth at low temperature)

(II) PARAMAGNETIC SUBSTANCES

These are the materials which when placed in a strong magnetic field acquire a feeble magnetism in the same sense as the applied magnetic field. The examples are platinum, aluminium, chromium, manganese, $CuSO_4, O_2$, air, etc.

Fig. 6.30 : Behaviour of Paramagnetic substance in an external field $ \vec{B} _0$

The characteristics of paramagnetic substances are

(a) They are attracted by a strong magnet

(b) Their susceptibility is positive but very small $(\chi>0)$

(c) Their relative permeability is slightly greater than unity. $(\mu>1)$

(d) Their susceptibility and permeability do not change with the variation of magnetising field.

(e) Their susceptibility is inversely proportional to temperature, $\bigg($ i.e. $.\chi \propto \frac{1}{T}\bigg)$

(f) Found in those material which have atoms containing odd number of electrons

(III) FERROMAGNETIC SUBSTANCES.

These are the substances which are strongly magnetised by relatively weak magnetising field in the same sense as the magnetising field. The exarmples are $Ni, Co$, iron and their alloys.

The characteristics of ferromagnetic substances are

(a) They are attracted even by a weak magnet.

(b) The susceptibility is very large and positive. $(\chi\gg0)$

(c) The relative permeability is very high (of the order of hundreds and thousands). ( $\mu \gg1$)

(d) The intensity of magnetisation is proportional to the magnetising field $H$ for smaller values, varies rapidly for moderate values and attains a constant value for larger values of $H$.

(e) The susceptibility of a ferromagnetic substance is inversely proportional to temperature i.e.,

$\chi \propto 1 / T \Rightarrow \chi=\frac{C}{T} ; C=$ curie constant.

This is called Curie law. At a temperature called curie temperature, ferromagnetic substance becomes paramagnetic. The curie temperatures for $Ni, Fe$ and $Co$ are $360^{\circ} C, 740^{\circ} C$ and $1100^{\circ} C$ respectively.

(f) Found in those material which have domains and can be converted into strong magnets

Keep in memory

HYSTERESIS

When a bar of ferromagnetic material is magnetised by a varying magnetic field and the intensity of magnetisation $I_m$ induced is measured for different values of magnetising field $H$, the graph of $I$ versus $H$ is as shown is fig. The graph shows

Fig. 6.33

(i) When magnetising field is increased from $O$ the intensity of magnetisation $I_m$ increases and becomes maximum (i.e. point $A$ ). This maximum value is called the saturation value.

(ii) When $H$ is reduced, $I_m$ reduces but is not zero when $H=0$. The remainder value $OB$ of magnetisation when $H=0$ is called the residual magnetism or retentivity. $OB$ is retentivity.

(iii) When magnetic field $H$ is reversed, the magnetisaiton decreases and for a particular value of $H$, it becomes zero i.e., for $H=$ $OC, I=0$. This value of $H$ is called the coercivity.

(iv) When field $H$ is further increased in reverse direction, the intensity of magnetisation attains saturation value in reverse direction (i.e., point D).

(v) When $H$ is decreased to zero and changed direction in steps, we get the part DFGA.

Properties of soft iron and steel

For soft iron, the susceptibility, permeability and retentivity are greater while coercivity and hysteresis loss per cycle are smaller than those of steel.

Permanent magnets are made of steel and cobalt while electromagnets are made of soft iron.

Oersted’s Experiment : An evidence of magnetic effect of current

After the invention of electric batteries, electric currents were a hot scientific topic, but no one suspected they had anything to do with magnetism. A Danish professor, Hans Christian Oersted, prepared for some friends, a science demonstration in his home. It included the heating of a wire by an electric current from a battery, and also some demonstrations of magnetism, using a compass on a stand. While carrying out the heating experiment, Oersted noted that every time he connected the current, the compass needle moved, too, something completely unexpected. No one else took notice. In the four months that followed he tried hard tc make sense of the phenomenon; but he couldn’t. The compass needle was neither attracted nor repelled, but tended to stand perpendicular to the wire.

Following Oersted’s article, the effect was confirmed by Andre-Marie Ampere, he felt that if a current in a wire exerted a magnetic force on a compass needle, two such wires also should interact magnetically. In a series of ingenious experiments he showed that this interaction was simple and fundamental, parallel (straight) currents attract, anti-parallel currents repel. The force between two long straight parallel currents was inversely proportional to the distance between them and proportional to the intensity of the current flowing in each. Loops of current attract or repel like magnets, coils with many loops multiplied the magnetic force, and Ampere guessed that iron atoms were magnetic because electric currents (soon named “Ampere currents”) circulated in them.

Fig. 6.35: Oersted’s experiment. Current in the wire deflects the compass needle.

CHECK POINT:

In figure, we see a magnet exerting a force on a current-carrying wire. Does a current-carrying wire exert a force on a magnet? Why or why n ?

Fig. 6.36

Show Answer

SOLUTION:

Yes, a current-carrying wire also exert a force on a magnet. The current-carrying wire produces a magnetic field around it. When a magnet is brought near the wire, its magnetic field interacts with the magnetic field created by the wire. Duc to this interaction the magnet experiences a force.

BIOT-SAVART LAW

The Biot-Savart Law defines the magnetic field $\vec{B}$ due to a current distribution in terms of the distribution. For example, let us consider a line distribution of steady current $I$ confined to a filamentary wire ST. The wire may be imagined to be subdivided into the many elementary parts, each characterised by a vector $d \vec{l}$ whose magnitude is equal to the length of elementary part and whose direction is that of the current, as shown.

The magnetic field at a point $P$ due to an elementary part is $d \vec{B}=(\frac{\mu_0 I}{4 \pi})(\frac{d \vec{l} \times \vec{R}}{R^{3}})$

where $\vec{R}$ is the position vector of point $P$ with respect to the elementary part. This magnetic field has the magnitude

$ d B=\bigg(\frac{\mu_0 I}{4 \pi}\bigg)\bigg(\frac{d l \sin \theta}{R^{2}}\bigg) $

and the direction of the cross product $(d \vec{l} \times \vec{R})$, i.e., inwards and perpendicular to the plane of figure, as shown. The magnetic field $\vec{B}$ at the point $P$ due to the complete wire ST is obtained by performing a summation process over all the elementary parts. The direction of $\vec{B}$ will be the same as that of an elementary magnetic field $d \vec{B}$.

Fig. 6.37

MAGNETIC FIELD DUE TO A CURRENT CARRYING CONDUCTOR

From Oersted experiment followed by Ampere we can conclude that a magnetic field is developed around a conductor when electric current is passed through it. This observation is called magnetic effect of electric current. The presence of a current in a wire near a magnetic compass affected the direction of the compass needle. We now know that current gives rise to magnetic fields, just as electric charge give rise to electric fields.

Fig. 6.38 : Compass near a current-carrying wire

Straight conductor

The magnetic field around a conductor carrying current is in the form of closed circular loops, in a plane perpendicular to theconductor, and is given by right hand thumb rule.

Right hand thumb rule

If we grasp the conductor in the palm of the right hand so that the thumb points in the direction of the flow of current, then the direction in which the fingers curl, gives the direction of magnetic field lines.

Fig. 6.39

For the current flowing through the conductor in the direction shown in figures (a) or (b), the rule predict that magnetic field lines will be in anticlockwise direction, when seen from above.

This rule is also called Maxwell’s corkscrew rule. If we consider ourselves driving a corkscrew in the direction of the current, then the direction of the corkscrew is the direction of the magnetic field.

Ampere’s swimming rule

Imagine a man swimming along the wire, in the direction of current, (such that the current enters at his feet and leaves him at his head) and facing towards a compass needle placed underneath the wire, then the magnetic field produced is such that the north pole of the compass needle gets deflected towards his left hand.

Knowledge ENHANCER

Magnetic filed due to long straight conducotr

$\mathrm{B}=\frac{\mu_0 I}{4 \pi r}\left(\sin \theta_1+\sin \theta_2\right) \quad \frac{\mu_0}{4 \pi}=10^{-7} \frac{\text { Tesla }-m}{\text { Amp. }}$

If the conductor is infinitely long then $\theta_1=90^\circ$ and $\theta_2=90^\circ$

$ B=\frac{\mu_0 I}{4 \pi r}\left[\sin \frac{\pi}{2}+\sin \left(\frac{\pi}{2}\right)\right]=\frac{\mu_0 I}{4 \pi r}[1+I] \frac{\mu_0 2 I}{4 \pi r} $

or $B=\frac{\mu_0 I}{2 \pi r}$

If conductor is of infinite length but one end is in front of point P i.e. one end of conductor starts from point N then

$\theta_1=0^{\circ}$ and $\theta_2=90^{\circ} \mathrm{B}=\frac{\mu_0 \mathrm{I}}{4 \pi \mathrm{r}}$

Conductor is finite length and point P is just in front of middle of the conductor

$\begin{array}{ll}B=\frac{\mu_0 I}{4 \pi r}(2 \sin \theta) & {\left[\theta_1=\theta_2=\theta\right]} \\ B=\frac{\mu_0 I L}{2 \pi r \sqrt{L^2+r^2}} & \sin \theta=\frac{\mathrm{L} / 2}{\sqrt{\left(\frac{L}{2}\right)^2+\mathrm{r}^2}}\end{array}$

CHECK POINT:

A beam of high-energy protons emerges from a cyclotron. Do you suppose there is a magnetic field associated with these particles? Why or why not?

Show Answer

SOLUTION:

Yes, I suppose there is a magnetic field associated with these particles. The particles are moving charged particles which produce a magnetic field.

ILLUSTRATION-6.4

Compute the magnetic field at a point of $9 cm$ from the long straight wire carrying a current of 6A.

Show Answer

SOLUTION:

$ \begin{aligned} & a=9 cm=9 \times 10^{-2} m, I=6 A \\ & B=\frac{\mu_o I}{2 \pi a}=\frac{4 \pi \times 10^{-7} \times 6}{2 \pi \times 9 \times 10^{-2}}=1.33 \times 10^{-5} T \end{aligned} $

CIRCULAR LOOP

At every point of a current-carrying circular loop, the concentric circles representing the magnetic field around it would become larger and larger as we move away from the wire (Fig. 6.42). By the time we reach at the centre of the circular loop, the arcs of these big circles would appear as straight lines. Every point on the wire carrying current would give rise to the magnetic field appearing as straight lines at the center of the loop. By applying the right hand rule, it is easy to check that every section of the wire contributes to the magnetic field lines in the same direction within the loop.

Knowledge ENHANCER

Magnetic field at the centre of a circular current-carrying coil

$ B=\frac{\mu_0 I}{2 r} $

Fig. 6.44

Magnetic dipole moment of the current loop

The current loop can be regarded as a magnetic dipole which produces its magnetic field and magnetic dipole moment of the current loop is equal to the product of ampere turns and area of current loop, we can write

$B=\frac{\mu_0 N I A}{2 r\left(\pi a^2\right)}=\frac{\mu_0}{2 \pi} \frac{N I A}{a^3}$ or $B=\frac{\mu_0}{4 \pi} \frac{2 N I A}{a^3} \quad \therefore \quad B=\frac{\mu_0}{4 \pi} \frac{2 M}{a^3}$

If the observation point is far away from the coil, then a « x S0. $a^2$ can be neglected in comparison to $x^2$.

$\therefore B=\frac{\mu_0 N \mathrm{I} a^2}{2 x^3}$

In terms of magnetic dipole moment $B=\frac{\mu_0}{4 \pi} \frac{2 M}{x^3} \quad\left[B=\frac{\mu_0}{2 \pi} \frac{N I A}{x^3}=\frac{\mu_0}{4 \pi} \frac{2 N \mathrm{I} A}{x^3}\right]$

RIGHT HAND PALM RULE

If we hold the thumb of right hand mutually perpendicular to the grip of the fingers such that the curvature of the finger represents the direction of current in the wire loop, then the thumb of the right hand will point in the direction of magnetic field near the centre of the current loop.

Fig. 6.45

ILLUSTRATION-6.5

An infinite straight conductor carrying current $2 I$ is split into a loop of radius $r$ as shown in figure. The magnetic field at the centre of the coil is

Fig. 6.46

Show Answer

SOLUTION:

Here, the wire does not produce any magnetic field at $O$ because the conductor lies on the line of $O$. Also, the loop does not produce magnetic field at $O$.

ILLUSTRATION-6.6

A current of $I$ ampere flows in a wire forming a circular arc of radius $r$ metres subtending an angle $\theta$ at the centre as shown. The magnetic field at the centre $O$ in tesla is

Fig. 6.47

Show Answer

SOLUTION:

$B=\frac{\mu_0 I}{2 r} \times \frac{\theta}{2 \pi}=\frac{\mu_0 I \theta}{4 \pi r}$

ILLUSTRATION-6.7

A conducting wire of infinite length, carrying a current $I$, is arranged in the shape, as shown. If the magnetic field at the centre $O$ of circular segment is zero, then, calculate the angle $\theta$.

Fig. 6.48

Show Answer

SOLUTION:

The magnetic field at the centre $O$ due to the two straight segments, taken together, is

$ B_1=2(\frac{\mu_0 I}{4 \pi}) $

which is perpendicular to the plane of figure and outwards, according to the ‘Right-Hand Rule’. The magnetic field at the centre $O$ due to the circular segment is

$ B_2=(\frac{\mu_0 I}{2 r})(\frac{2 \pi-\theta}{2 \pi}) $

which is perpendicular to the plane of figure and inwards, according to the ‘Modified Right-Hand Rule’. Then, the resultant magnetic field at the centre $O$ will be zero, when

$ \begin{aligned} B_1 & =B_2 \\ \Rightarrow 2\biggl(\frac{\mu_0 I}{4 \pi r}\biggl) & =\biggl(\frac{\mu_0 I}{2 r}\biggl)\biggl(\frac{2 \pi-\theta}{2 \pi}\biggl) \\ \Rightarrow \theta & =2(\pi-1) \end{aligned} $

ILLUSTRATION-6.8

A long conducting wire, carrying a current $I$, is arranged in the shape, as shown, with the two straight segments in the plane of figure and the circular loop of radius $r$ perpendicular to the plane of figure. Calculate the magnetic fies circular loop.

Fig. 6.49

Show Answer

SOLUTION:

The magnetic field at the centre $O$ of the circular loop due to the two straight segments, taken together, will be

$ B_1=2\bigg(\frac{\mu_0 I}{4 \pi r}\bigg) $

which is perpendicular to the plane of figure and outwards, according to the ‘Right-Hand Rule’. The magnetic field at the centre $O$ due to the circular loop will be

$ B_2=\bigg(\frac{\mu_0 I}{2 r}\bigg) $

which is parallel to the straight segments and towards right. Since, the two magnetic fields $B_1, B_2$ are mutually perpendicular, the resultant magnetic field at the centre $O$ is given by

$ B=\sqrt{B_1^{2}+B_2^{2}}=(\frac{\mu_0 I}{2 r}) \sqrt{(\frac{1}{\pi^{2}}+1)} $

FORCE ON A CHARGED PARTICLE IN A MAGNETIC FIELD

According to Lorentz, charge particle $q$ moving with velocity $v$ in a magnetic field $B$ experiences

Force, $\quad$ $\vec{F}=q(\vec{v} \times \vec{B})$

The force $\vec{F}$ is always perpendicular to both the velocity $\vec{v}$ and the field $\vec{B}$.

A charged particle at rest in a steady magnetic field does not experience any force.

If the charged particle is at rest then $\vec{v}=0$, so $\vec{v} \times \vec{B}=0$

A moving charged particle does not experience any force in a magnetic field if its motion is parallel or antiparallel to the field. i.e., if $\theta=0^{\circ}$ or $180^{\circ}$,

then, $\quad$ $|\vec{F}|=q v B \sin \theta=0 \quad\left[\because \sin 0^{\circ}=0\right.$ and $\left.\sin 180^{\circ}=0\right]$

Fig. 6.50

If the particle is moving perpendicular to the field.

In this situation all the three vectors $\vec{F}, \vec{v}$ and $\vec{B}$ are mutually perpendicular to each other.

Then $\sin \theta=\max =1$, i.e., $\theta=90^{\circ}$,

The force will be maximum $F _{\max }=q v B$

Work done by force due to magnetic field in motion of a charged particle is always zero.

When a charged particle move in a magnetic field, then force acts on it is always perpendicular to displacement,

so the work done, $\quad W=\overrightarrow{F} . \overrightarrow{s}=Fs \cos \theta=0 \quad($ as $\theta=90^{\circ}$),

And as by work-energy theorem $W=\Delta KE$, the kinetic energy $(=\frac{1}{2} mv^{2})$ remains unchanged and hence speed of charged particle $v$ remains constant.

However, in this situation the force changes the direction of motion, so the direction of velocity $\vec{v}$ of the charged particle changes continuously.

The direction of the cross product can be obtained by using a right-hand rule:

Fingers of the right hand point in the direction of the first vector $(v)$ in the cross product, then adjust your wrist so that you can bend your fingers (at the knuckles) toward the direction of the second vector (B); extend the thumb to get the direction of the force.

Fig. 6.51 : Magnetic force acting on a moving charge

A charged particle moving in a plane perpendicular to a magnetic field will move in a circular orbit with the magnetic force playing the role of centripetal force. The direction of the force is given by the right-hand rule. Equating the centripetal force with the magnetic force and solving for $R$ the radius of the circular path we get,

$m v^{2} / R=q v B$ and $R=m v / q B$

Fig. 6.52 : Orbit of charged particle in a magnetic field.

If charge particle projected at some angle with magnetic field it will move on helical path as shown.

Fig. 6.53 : Helical path.

ILLUSTRATION-6.9

A magnetic field of $(4.0 \times 10^{-3} \hat{k}) T$ exerts a force of $(4.0 \hat{i}+3.0 \hat{j} \times 10^{-10}) N$ on a particle, having a charge of $1.0 \times 10^{-9} C$ and moving in the $x-y$ plane. Find the velocity of particle.

Show Answer

SOLUTION:

As the particle is moving in the $x-y$ plane, it’s velocity can be expressed as $\vec{v}=(v_x \hat{i}+v_y \hat{j}) m / s$. Then, the magnetic force on it becomes

$ \begin{aligned} \vec{F} _m & =q(\vec{v} \times \vec{B}) \\ & =1.0 \times 10^{-9}[(v_x \hat{i}+v_y \hat{j}) \times(4.0 \times 10^{-3} \hat{k})] \\ & =+4.0 \times 10^{-12} v_y \hat{i}-4.0 \times 10^{-12} v_x \hat{j} \equiv(4.0 \times \hat{i}+3.0 \hat{j}) 10^{-10} \end{aligned} $

Comparing the coefficients of unit vectors of both sides, we have

$ v_x=-75, v_y=+100 $

and then, the velocity of particle is given by

$ \vec{v}=(-75 \hat{i}+100 \hat{j}) m / s $

ILLUSTRATION-6.10

A charge of $2.0 \times 10^{-6} C$ is moving with a speed of $2.0 \times 10^{6} m / s$ in the $+x$ direction in a magnetic field $\vec{B}=(0.2 \hat{j}+0.4 \hat{k}) T$.

Find the magnetic force on the charge.

Show Answer

SOLUTION:

As the charge is moving with a speed of $2.0 \times 10^{-6} m / s$ in the $+x$ direction, it’s velocity can be expressed as

$ \vec{v}=(2.0 \times 10^{6}) \hat{i} $ m/s

and then, the magnetic force on the charge will be

$ \begin{aligned} \vec{F} _m & =q(\vec{v} \times \vec{B}) \\ & =2.0 \times 10^{-6}[(2.0 \times 10^{6} \hat{i}) \times(0.2 \hat{j}+0.4 \hat{k})] \\ & =4.0[\hat{i} \times(0.2 \hat{j}+0.4 \hat{k})] \\ & =0.8(\hat{i} \times \hat{j})+1.6(\hat{i} \times \hat{k}) \\ & =(-1.6 \hat{j}+0.8 \hat{k}) N \end{aligned} $

Knowledge ENHANCER

VELOCITY SELECTOR :

Fig. 6.65 : Velocity selector for charged particles

An electric field and a magnetic field placed at right angles to each other can function as a “velocity selector.” When force up = force down, the charge will travel in a straight (horizontal) line. The speed can be obtained from the equation

$ q v B=q E, \text{ or } v=E / B $

Charged particles leaving a velocity selector (with a known velocity) can be inserted into a chamber with a magnetic field as shown.

In the circular orbit equation above $R=m v / q B$.

We can substitute $v=E / B$ to get $R=m E / q B^{2}$

from which we can solve for $m / v$, the mass-to-charge ratio. Knowing the charge (ionized state) and the measured radius we can find the mass of the particle.

Fig. 6.54 : Mass spectrometer.

Force on a current carrying conductor in a magnetic field

Similar to the force on a moving charge in a $B$ field, we have for a conductor of length $L$ carrying a current of $I$ in a $B$ field the force experienced by the conductor: $\vec{F}=I \vec{L} \times \vec{B}$

Fig. 6.55 : Force on a moving charge in a current-carrying conductor.

Direction of force can be determined by fleming’s left hand rule, right hand palm rule or screw rule.

Fleming’s Left-hand Rules

Stretch the fore-finger, central finger and thumb of left hand mutually perpendicular.

Fig. 6.56

Then if the fore-finger points in the direction of field $(\vec{B})$, the central finger in the direction of current $I$, the thumb will point in the direction of force.

(ii) Right-hand Palm Rule

Stretch the fingers and thumb of right hand at right angles to each other. ihen if the fingers point in the direction of field $(\vec{B})$ and thumb in the direction of current $I$, the normal to palm will point in the direction of force.

Fig. 6.57

To understand it in better manner see three figures.

Fig. 6.58

If no current is flowing through the wire obviously the force will be zero, if current is flowing upward using right hand rule( field inward shown by x) you will find force will act in left direction, hence wire will deflect as shown and if direction of current is downward force will try to move the wire in right direction and will deflect the wire as shown. Screw rule: Curl right hand finger from current towards magnetic field thumb will direction of force as shown in figure.

Fig. 6.59 : Magnetic force on straight wire segment.

You can see component of magnetic field perpendicular to current contributes in force on conductor. If current in wire is along external magnetic field force will be zero and if perpendicular to field it will be maximum.

Fig. 6.60

If direction of current or field reverses direction of force also reverses as shown in figure 6.61

Fig. 6.61 : Magnetic forces - reversing the direction of the B field or the direction of the current I reverses the direction of the force.

Ay arbitrary shape in a uniform field experience a force

$ \overrightarrow{F}=I \overrightarrow{\ell^{\prime}} \times \overrightarrow{B} $

where $\vec{\ell}^{\prime}$ is the length vector joining initial and final points of the conductor as shown in figure.

If the current-carrying conductor in the form of a loop of any arbitrary shape is placed in a uniform field, then, $\overrightarrow{F}=0$ i.e., the net magnetic force on a current loop in a uniform magnetic field is always zero as shown in figure 6.62 (a).

Here it must be kept in mind that in this situation different parts of the loop may experience elemental force due to which the loop may be under tension or may experience a torque as shown in figure 6.62 (b).

Fig. 6.62

CURRENT LOOP IN A UNIFORM FIELD

If a current-carrying conductor is situated in a non-uniform field, its different elements will experience different forces; so in this situation,

$\vec{F}_{\mathrm{R}} \neq 0$ but $\vec{\tau}$ may or may not be zero

Fig. 6.63

ILLUSTRATION-6.11

A vertical straight conductor $X$ of length $0.5 m$ is situated in a uniform horizontal magnetic field, of $0.1 T$ (i) calculate the force on $X$ when a current of $4 A$ is passed into it (ii) through what angle must $X$ be turned in a vertical plane so that the force on $X$ is halved?

Show Answer

SOLUTION:

(i) Here L = 0.5 m, B = 0.1 T, I = 4 A, F = ?

F = BIL = $0.1\times 4\times 0.5=0.2N$

(ii) Let $\alpha$ be the required angle.

The force on the conductor

$\begin{aligned} F^{\prime} & =\frac{1}{2} F \frac{0.2}{2}=0.1 \mathrm{~N} \\ F^{\prime} & =B I L \operatorname{Sin} \theta\end{aligned}$

$\begin{aligned} & \text { or } \frac{1}{2} \mathrm{BIL}=\mathrm{BILSin}\left(90^{\circ}-\alpha\right) \\ & \text { or } \frac{1}{2} \cos \alpha \text { or } \alpha=\cos ^{-1}\left(\frac{1}{2}\right)=60^{\circ}\end{aligned}$

Fig. 6.64

ILLUSTRATION-6.12

A rectangular loop of sides $25 cm$ and $10 cm$ carrying a current of $15 A$ is placed with its longer side parallel to a long straight conductor $2.0 cm$ apart carrying a current of $25 A$. What is the net force on the loop?

Show Answer

SOLUTION:

Consider a rectangular loop PQRS placed near a long straight conductor AB as shown in Figure 6.65. Due to the interaction of currents, the arm PQ of the loop will get attracted while arm RS will get repelled. Forces on the arms QR and SP will be equal and opposite and hence cancel out.

Fig. 6.65

Here, $\quad PQ=25 cm=25 \times 10^{-2} m$,

$PS=10 cm=10 \times 10^{-2} m$

Distance of $PQ$ from $AB$

$ r_1=2.0 cm=2.0 \times 10^{-2} m $

Distance of RS from $AB$,

$ r_2=2.0+10=12.0 cm=12.0 \times 10^{-2} m $

Current through long wire $A B, I_1=25 A$

Current through rectangular loop, $I_2=15 A$

$\therefore$ Force on the arm $PQ, F_1=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r_1} \times$ length $P Q$

$F_1=\frac{10^{-1} \times 2 \times 25 \times 15 \times 25 \times 10^{-2}}{2.0 \times 10^{-2}}$

$=9.375 \times 10^{-4}$ $\quad$ (towards $AB$)

Force on the arm RS,

$F_2=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r_2} \times$ length $RS=\frac{10^{-7} \times 2 \times 25 \times 15 \times 25 \times 10^{-2}}{12 \times 10^{-2}}$

$=1.563 \times 10^{-4}$ (away from AB)

$\therefore$ Effective force on the loop,

$F=F_1-F_2=9.375 \times 10^{-4}-1.563 \times 10^{-4}=7.812 \times 10^{4} N$

(towards AB)

FORCE BETWEEN CURRENT CARRYING WIRES

If currents are in the same direction, wires attract. If in opposite direction, wires repel. This is the opposite of the rule for charges: like charges repel, opposite charges attract. Using the two right hand rules, one for finding the direction of the B-field of a wire, and the other for the direction of force on a wire, you can predict the results.

Fig. 6.66

The force per unit length on one wire due to other wire

$$ \frac{F}{\ell}=\frac{\mu_0 I_1 I_2}{2 \pi d} $$

[Where $I_1, I_2$ current in two wires and $d=$ distance between wire]

Experimental Demonstration

Definition of Ampere

$\because \quad \mathrm{F}=\frac{\mu_0}{2 \pi} \frac{I_1 I_2}{r} \mathrm{~N} / \mathrm{m}$

When $I_1=I_2=1$ ampere and $r=1 m$, then $F=\frac{\mu_0}{2 \pi}=\frac{4 \pi \times 10^{-7}}{2 \pi} N / m=2 \times 10^{-7} N / m$

This leads to the following definition of ampere.

One ampere is that current which, if passed in each of two parallel conductors of infinite length and one metre apart in vacuum causes each conductor to experience a force of $2 \times 10^{-7}$ newton per metre of length of conductor.

ILLUSTRATION-6.13

Calculate the magnetic force per unit length experienced by a long straight wire $W_1$ carrying a current of $2 A$ and placed parallel to and exactly midway between the two long, straight, and parallel wires $W_2$ and $W_3$ carrying the currents of $2 A$ and $4 A$, respectively, in the opposite directions and separated by $0.1 m$.

Fig. 6.68

Show Answer

SOLUTION:

The wire $W_1$ experiences an attractive magnetic force $F _{12}$ per unit lenght towards left due to the wire $W_2$, carrying the parallel current.

$ F _{12}=\frac{\mu_0 \times 2 \times 2 \times 1}{2 \pi \times 0.05}=1.6 \times 10^{-5} Nm^{-1} $

The wire $W_1$ also experiences a repulsive magnetic force $F _{13}$ per unit length towards left due to the wire $W_3$, carrying the antiparallel current.

$ F _{13}=\frac{\mu_0 \times 2 \times 4 \times 1}{2 \pi \times 0.05}=3.2 \times 10^{-5} Nm^{-1} $

Since, both magnetic forces on the wire $W_1$ are in the same direction towards left, the resultant magnetic force per unit length on it becomes

$ F=F _{12}+F _{13}=4.8 \times 10^{-5} Nm^{-1} $

ELECTROMAGNET

A magnetic field is produced when an electric current flows through a coil of wire. This is the basis of the electromagnet. We can make an electromagnet stronger by doing these things:

wrapping the coil around an iron core

adding more turns to the coil

increasing the current flowing through the coil.

The magnetic field around an electromagnet is just the same as the one around a bar magnet. It can, however, be reversed by turning the battery around. Unlike bar magnets, which are permanent magnets, the magnetism of electromagnets can be turned on and off just by closing or opening the switch.

Uses of electromagnet

1. For lifting and transporting large masses of iron scrap, girders, plates etc., especially to places where it is not convenient to take the help of human labour. Electromagnets are used to lift as much as 20-22 tonnes of iron in a single lift. To unload the magnet at the desired place, the current in the electromagnet is switched off so that the load drops. 2. For loading furnaces with iron. 3. For separating magnetic substances such as iron from other debris (e.g. for separating iron from the crushed copper ore in copper mines). 4. For removing pieces of iron from wounds. 5. In several electrical devices such as electric-well, telegraph, electric tram, electric motor, relay, microphone, load speaker, etc. 6. In scientific research to study the magnetic properties of a substance in a magnetic field.

Fig. 6.70 : Electric Crane.

Some useful devices based on electromagnets produced by currents passing through solenoids

Electric bell

Parts of an electric bell

P - Push button, E - Electromagnet, A - Soft iron armature, S - Flexible metal strip, C - Contact screw, H - Hammer, G - Gong

When the push button $P$ is depressed, current will flow through the circuit, and this will energise the electromagnet, E. It will attract the soft iron armature, A, causing the hammer to strike the gong. Contact at the screw $C$, will be broken, de-energising the electromagnet. The flexible metal strip, $S$ acts as a spring, pulling back the armature and the hammer, and restoring contact at the screw. This cycle is repeated as long as the push button is depressed.

Fig. 6.71

Telephone

A typical telephone handset consists of two parts - the receiver (that you place next to your ear) utilises an electromagnet, and a microphone, into which you speak.

Fig. 6.72

The receiver has an electromagnet, B, connected to the circuit. Varying strengths of the current cause different amount of magnetisation of the electromagnet. The electromagnet causes deformation of the steel membrane, A. This results in sound waves that are picked up by the user’s ear.

Fig. 6.73

The microphone consists of a steel membrane, D, lying across a small container of carbon powder, C. Two electrical contacts, E are embedded in the carbon powder, whose resistance depends on the density of the powder particles. Sound waves compress the powder to varying degrees, thus altering the resistance of the carbon powder. This in turn causes small changes in the current flowing through the circuit.

Loudspeaker

Fig. 6.74

A loudspeaker consists of a central pole, surrounded by a ring magnet, A movable coil, is wound around the central pole. Attached to this coil, which carries the current, is a cardboard cone.

When a current passes through the coil, a force acts on the coil forcing it to move in and out. The paper cone which is attached Relays to it moves with it, and sets up sound waves, which reproduce the sounds which a microphone originally converted to a current.

Relays

A relay is a device whereby a secondary circuit can be switched on by a primary circuit.

Referring to the diagram above, when the switch $S$ is closed, the iron core inside the solenoid becomes magnetised. It attracts the armature, which is pivoted. The upper limb of the armature causes the contacts in the secondary circuit $XY$ to close, thus activating that circuit.

Fig. 6.75

DC MOTOR

A d.c. motor converts direct current energy from a battery into mechanical energy of rotation.

Principle :

It is based on the fact that when a coil carrying current is held in a magnetic field, it experiences a torque, which rotates the coil.

Construction :

It consists of the five parts.

Armature :

The armature coil ABCD consists of a large number of turns of insulated copper wire wound over a soft iron core, Field Magnet : The magnetic field is supplied by a permanent magnet $NS$.

Split-rings or Commutator :

These are two halves of the same ring. The ends of the armature coil are connected to these halves which also rotate with the armature.

Brushes :

These are two flexible metal plates or carbon rods $B_1$ and $B_2$, which are so fixed that they constantly touch the revolving rings.

Battery :

The battery consists of a few cells of voltage $V$ connected across the brushes. The brushes convey the current to the rings, from where it is carried to the armature.

Working :

The battery sends current through the armature coil in the direction shown in figure. Applying Fleming’s Left Hand Rule, $CD$ experiences a force directed inwards and perpendicular to the plane of the coil. Similarly, AB experiences a force directed outwards and perpendicular to the plane of the coil. These two forces being equal, unlike and parallel form a couple. The couple rotates the armature coil in the anticlockwise direction. After the coil has rotated through $180^{\circ}$, the direction of the current in $AB$ and $CD$ is reversed, figure. Now $CD$ experiences an outward force and $AB$ experiences an inward force. The armature coil thus continues rotating in the same i.e., anticlockwise direction.

Fig. 6.76

Efficiency of the d.c. motor :

Since the current $I$ is being supplied to the armature coil by the external source of e.m.f. V, therefore, Input electric power $=VI$

According to Joule’s law of heating, Power lost in the form of heat in the coil $=I^{2} R$

If we assume that there is no other loss of power, then Power converted into external work i.e.,

Output mechanical power $=V I-I^{2} R=(V-I R) I=E I$

$\therefore$ Efficiency of the d.c. motor, $\eta=\frac{\text{ Output mechanical power }}{\text{ Input electric power }}$

A d.c. motor delivering maximum output has an efficiency of only $50 \%$.

Uses :

(1) The d.c. motors are used in d.c. fans (exhaust, ceiling or table) for cooling and ventilation.

(2) They are used for pumping water.

(3) Big d.c. motors are used for running tram-cars and even trains.

AC Motor

As in the DC motor case, a current is passed through the coil, generating a torque on the coil. Since the current is alternating, the motor will run smoothly only at the frequency of the sine wave. It is called a synchronous motor. More common is the induction motor, where electric current is induced in the rotating coils rather than supplied to them directly.

One of the drawbacks of this kind of AC motor is the high current which must flow through the rotating contacts. Sparking and heating at those contacts can waste energy and shorten the lifetime of the motor. In common AC motors the magnetic field is produced by an electromagnet powered by the same AC voltage as the motor coil. The coils which produce the magnetic field are sometimes referred to as the “stator”, while the coils and the solid core which rotates is called the “armature”. In an AC motor the magnetic field is sinusoidally varying, just as the current in the coil varies.

Fig. 6.77

Galvanometer

The torque on a current loop in a uniform magnetic field is used to measure electrical magnetic field is used to measure electrical currents. This current measuring device is called a moving coil galvanometer.

Fig. 6.78

The galvanometer consists of a coil of wire often rectangular, carrying the current to be measured. There are generally many turns in the coil to increase its sensitivity. The coil is placed in a magnetic field such that the lines of B remain nearly parallel to the plane of wire as it turns. This is achieved by having a soft iron cylinder placed at the center of the coil. Magnetic field lines tend to pass through the iron cylinder, producing the field configuration. The moving coil is hung from a spring which winds up as the coil rotates; this winding up produces a restoring torque proportional to the winding up (or twisting) of the spring, i.e. to the angular deflection of the coil. The coil comes to equilibrium when this restoring torque $k$ balances the torque due to the magnetic field balances the torque due to the magnetic field.

CHECK POINT:

How could a light bulb near, yet not touching, an electromagnet be lit? Is AC or DC required? Defend your answer.

Show Answer

SOLUTION:

For the bulb to be lit, $AC$ should be supplied in the coil of the electromagnet so that an alternating magnetic field will be established in the region around it. This changing magnetic field will induce a current in the circuit of the light bulb and the bulb will glow.

MISCELLANEOUS

SOLVED EXAMPLES

1. Two identical wires $R$ and $S$ lie parallel in the horizontal plane, their axis being 0.10 m apart. A current of 30 A in $S$ Neglecting the effect of the earth’s magnetic fux density at a point $P$ in the plane of the wires. If $P$ is (i) mid way between $R$ and $S$ (ii) 0.05 $m$ from $R$ and $0.15 m$ from $S$. [the permeability of free space $\mu_0=4 \pi \times 10^{-7} Hm^{-1}$ ]

Show Answer

SOLUTION: (i) Midway between $R$ and $S$

Field due to $R, B_R=\frac{\mu_o I}{2 \pi r}$

Field due to $S, \quad B_s=\frac{\mu o I}{2 \pi r}$

$\frac{4 \pi \times 10^{-7} \times 30}{2 \pi \times 0.05}=12 \times 10^{-5} \mathrm{~T}$

Direction of $B$ due to two wire is the same, total flux density,

$B=B_R+B_S$

$=4 \times 10^{-5}+12 \times 10^{-5}=16 \times 10^{-5}=1.6 \times 10^{-4} T$

(ii) If $P$ is $0.05 m$ from $R$ and $0.15 m$ from $S$

$ \begin{aligned} & B_R=\frac{\mu_o I}{2 \pi r}=\frac{4 \pi \times 10^{-7}}{2 \pi \times 0.05}=4 \times 10^{-5} T \\ & B_S=\frac{\mu_0 I}{2 \pi r}=\frac{4 \pi \times 10^{-7} \times 30}{2 \pi \times 0.15} \end{aligned} $

$\because \quad B_R$ and $B_S$ are in opposite direction

Resultant field $=B_S-B_R=4 \times 10^{-5}-4 \times 10^{-5}=0$

2. In the figure $X$ is a very long straight wire.Rectangle PQRS is suspended with PS 2 cm from $x$ as shown the dimensions of PQRS are 10 cm by 3 cm, and current of 2A flows in the coil. Calculate the resultant force on PQRS in magnitude and direction.

Show Answer

SOLUTION: The force on the side PS due to current $X$,

$F= Bil$

$=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\mathrm{i}_{\mathrm{x}}}{\mathrm{r}} \times \mathrm{i} \ell=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{5}{2 \times 10^{-2}} \times 0.10$

$=10 \times 10^{-6} N$ along + ve $x-$ axis

The force on the side QR due to current through $X$,

$F=Bil$

$=\frac{\mu_{\mathrm{o}}}{2 \pi} \frac{\mathrm{i}_{\mathrm{x}}}{\mathrm{r}} \times \mathrm{i} \ell=\frac{4 \pi \times 10^{-7}}{2 \pi} \times \frac{5}{5 \times 10^{-2}} \times 0.10$

$=4 \times 10^{-6} N$ along -ve $x$-axis

The force on $PQ$ and $SR$ due to the current carrying conductor are equal and opposite.

Hence the net force $=10 \times 10^{-6}-4 \times 10^{-6}$

$=6 \times 10^{-6} N \text{ along }$ +ve $x$-axis

3. Calculate the magnetic flux density at a point $20 cm$ to the west of a long vertical wire carrying a current of $1.0 A$ flowing up the wire. [the horizontal component of the earth’s field at the point is $1.8 \times 10^{-5}$]

Show Answer

SOLUTION: $I=1.0 A, r=20 cm=0.2 m$

$ B=\frac{\mu_0 I}{2 \pi r}=\frac{4 \pi \times 10^{-7} \times 1.0}{2 \pi \times 0.2}=1.0 \times 10^{-6} T=10 \times 10^{-5} T $

From the right hand grip rule, the direction of the field is from north to south and the horizontal component of the earth’s field is from south to north.

Hence the resulting magnectic flux density

$=10 \times 10^{-5} T-1.8 \times 10^{-5} T$

$=8.2 \times 10^{-5} T$ $\quad$ (From north to south)

4. A horizontal straight wire $5 cm$ long weighing $1.2 gm^{-1}$ is placed perpendicular to a uniform horizontal magnetic field of the flux density $0.6 T$. If the resistance of the wire is $3.8 \Omega m^{-1}$ calculate the potential difference that has to be applied between the ends of the wire to make it just supporting $(g=9.8 ms^{-1})$

Show Answer

SOLUTION: $1=5 cm=5 \times 10^{-2} m$

Mass of the wire,

$m=$ mass per unit length $\times$ length of the wire.

$=1.2 \times 10^{-3} kg m^{-1} \times 5 \times 10^{-2} m=6 \times 10^{-5} Kg$,

$B=0.6 T$

Resistance of the wire

R = (Resistance per unit length $x$ length of the wire)

$=3.8 \times 5 \times 10^{-2}=019 \Omega$

Let $V$ be the P.d. applied between the ends of the wire, for the wire to be self supporting:

$mg=BIl=B \frac{V}{R} l$

$\therefore \quad V=\frac{mg \times R}{B \ell}=\frac{6 \times 10^{-5} \times 9.8 \times 0.19}{0.6 \times 5 \times 10^{-2}}=3.72 \times 10^{-3} V$

5. Two long parallel conductors carry respectively currents of 12 and 8A respectively in the same direction. If the wires are $10 cm$ apart, find where the third parallel wire also carrying a current must be placed so that the force experienced by it shall be zero.

Show Answer

SOLUTION: For the force on the third conductor to be zero, the direction of the flux density due to the current flowing in the two wires must be opposite in the position of the wire.

$\therefore$ The third wire must be placed between the wire. Let the third wire placed at a distance $x m$ from the wire carrying 12A, then,

$B_1=B_2$

$\frac{\mu_o I_1}{2 \pi x}=\frac{\mu_o I_2}{2 \pi(0.1-x)}$

or $\frac{12}{x}=\frac{8}{0.1-x}$ or $\frac{3}{x}=\frac{2}{0.1-x}$

or $0.3=5 x=\frac{0.3}{5}=0.06 m$

<img src=“https://cdn.mathpix.com/cropped/2024_02_05_937ddbfbb920f311aae4g-073.jpg?height=165&width=263&top_left_y=2005&top_left_x=882)

6. A rectangular loop of sides $25 cm$ and $10 cm$ carrying a current of $15 A$ is placed with its longer side parallel to a long straight conductor $2.0 cm$ apart carrying a current of $25 A$. What is the net force on the loop?

Show Answer

SOLUTION: Consider a rectangular loop PQRS placed near a long straight conductor $AB$ as shown in figure. Due to the interaction of currents, the arm PQ of the loop will get attracted while arm RS will get repelled. Forces on the arms QR and SP will be equal and opposite and hence cancel out.

Here, $\quad P Q=25 cm=25 \times 10^{-2} m$,

$ P S=10 cm=10 \times 10^{-2} m $

Distance of PQ from $A B$,

$ r_1=2.0 cm=2.0 \times 10^{-2} m $

Distance of RS from $A B$,

$ r_2=2.0+10=12.0 cm=12.0 \times 10^{-2} m $

Current through long wire $AB, I_1=25 A$

Current through rectangular loop, $I_2=15 A$

$\therefore$ Force on the arm $P Q, F_1=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r_1} \times$ length $P Q$

<img src=“https://cdn.mathpix.com/cropped/2024_02_05_937ddbfbb920f311aae4g-074.jpg?height=349&width=294&top_left_y=722&top_left_x=1095)

$F_1=\frac{10^{-1} \times 2 \times 25 \times 15 \times 25 \times 10^{-2}}{2.0 \times 10^{-2}}$

$=9.375 \times 10^{-4}$ $\quad$ (towards $AB$ )

Force on the arm RS,

$F_2=\frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r_2} \times$ length $RS=\frac{10^{-7} \times 2 \times 25 \times 15 \times 25 \times 10^{-2}}{12 \times 10^{-2}}$

$ =1.563 \times 10^{-4} \text{ (away from AB) } $

$\therefore$ Effective force on the loop,

$F=F_1-F_2=9.375 \times 10^{-4}-1.563 \times 10^{-4}=7.812 \times 10^{4} N \quad$ (towards $AB$ )

7. A narrow vertical rectangular coil is suspended from the middle of its upper side with its plane parallel to a uniform horizontal magnetic field of $0.02 T$. The coil has 10 turns and the lengths of its vertical and horizontal sides are $0.1 m$ and $0.05 m$ respectively. Calculate the torque on the coil when a current of $5 A$ is passed into it. What would be the new value of the torque if the plane of the vertical coil was initially at $60^{\circ}$ to the magnetic field and a current of $5 A$ was passed into the coil.

Show Answer

SOLUTION: $B=0.02 T, N=10$ Turns

$A=1 \times b=0.1 \times 0.05=0.005 m^{2}$

$I=5 A$

Torque $=$ BINA $=0.02 \times 5 \times 10 \times 0.005$

$ =0.005 Nm=5 \times 10^{-3} Nm $

New value of the torque when the plane of the vertical coil was at $60^{\circ}$ to the magnetic field.

$ =BINA \cos \theta=5 \times 10^{-3} \cos 60^{\circ} $

$ =5 \times 10^{-3} \times 1 / 2=2.5 \times 10^{-3} Nm . $

Fill in the Blanks:

DIRECTIONS : Complete the following statements with an appropriate word / term to be filled in the blank space(s).

1. A compass needle is a …………….. magnet.

Show Answer

Answer: small

2. Field lines are used to represent a ……………..

Show Answer

Answer: magnetic field

3. Field lines are shown closer together where the magnetic field is ……………..

Show Answer

Answer: greater

4. A metallic wire carrying an electric current has associated with it a …………….. field.

Show Answer

Answer: magnetic

5. The field lines about the wire consist of a series of concentric circles whose direction is given by the …………….. rule.

Show Answer

Answer: right-hand

6. The magnetic lines of force are the lines drawn in a magnetic field along which a …………….. pole would move.

Show Answer

Answer: north magnetic

7. An electric current can be used for making temporary magnets known as ……………..

Show Answer

Answer: electromagnets

8. The unit of magnetic field is ……………..

Show Answer

Answer: tesla

9. The S.I. unit of magnetic flux ……………..

Show Answer

Answer: Weber

10. The force between currents is called the …………….. force.

Show Answer

Answer: magnetic

11. The N-pole of a compass points to the …………….. pole of a permanent magnet.

Show Answer

Answer: S

12. The force that a magnetic field exerts on a current is always perpendicular to the …………….. and to the ……………..

Show Answer

Answer: field, current

13. In a magnetic field pointing away from you, an electron traveling to the right will experience a force in the …………….. direction.

Show Answer

Answer: downward

14. Magnetic fields are produced by ……………..

Show Answer

Answer: currents

15. You are looking into a solenoid, at its S-pole, along its axis From your view point, the direction of the current in the solenoid is ……………..

Show Answer

Answer: clockwise

16. Crowding the wires of a solenoid more closely together will …………….. the strength of the field inside it.

Show Answer

Answer: increase

17. A paramagnet magnet behaves like a solenoid because both contain currents in the form of ……………..

Show Answer

Answer: circles

18. Magnetic field lines emerge from the …………….. pole of a solenoid or a permanent magnet.

Show Answer

Answer: None

19. You are looking down the axis of a solenoid, and the current from your position is clockwise. The end of the solenoid facing you is a …………….. pole.

Show Answer

Answer: south

True / False :

DIRECTIONS : Read the following statements and write your answer as true or false.

1. Energy associated with an electric field is analogous to potential energy whereas the energy associated with the magnetic field is analogous to kinetic energy.

Show Answer

Answer: True

2. No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field.

Show Answer

Answer: True

3. An electron and a proton move in a uniform magnetic field with same speed perpendicular to the magnetic field. They experience forces in opposite directions differing by a factor of 1840.

Show Answer

Answer: False

4. A positive charge projected along the axis of a current carrying solenoid moves undeviated from its original path.

Show Answer

Answer: True

5. There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it.

Show Answer

Answer: False

6. An electron does not suffer any deflections while passing through a region. This makes sure that there is no magnetic field in that region.

Show Answer

Answer: True

7. The field at the centre of a long circular coil carrying current will be parallel straight lines.

Show Answer

Answer: True

8. A magnetic field exists in the region surrounding a magnet, in which the force of the magnet can be detected.

Show Answer

Answer: True

9. The pattern of the magnetic field around a conductor due to an electric current flowing through it depends on the shape of the conductor.

Show Answer

Answer: True

10. A current-carrying conductor when placed in a magnetic field always experiences a force.

Show Answer

Answer: False

11. ‘Lodestone’ is a naturally occurring magnet.

Show Answer

Answer: True

12. The direction of force on a current carrying conductor placed in a magnetic field can be reversed by reversing the direction of current flowing in the conductor.

Show Answer

Answer: True

13. The direction of force on a current carrying conductor placed in a magnetic field cannot be reversed by reversing the direction of magnetic field.

Show Answer

Answer: False

14. Two magnetic lines of force never intersect each other.

Show Answer

Answer: True

15. The field lines inside the infinite solenoid are in the form of parallel straight lines.

Show Answer

Answer: True

Match the Following :

DIRECTIONS : Each question contains statements given in two columns which have to be matched. Statements $(A, B, C, D)$ in column I have to be matched with statements $(p, q, r, s)$ in column II.

1.

	Column I		Column II
(A)	Work done by the magnetic force may be	(p)	zero
(B)	Work done by the pseudo force may be	(q)	$\pm$ ve, zero
(C)	Frictional work	(r)	Conservative
(D)	Change in kinetic energy of charge particle in magnetic field.	(s)	Non conservative

Show Answer

Answer: $(A) \to \mathbf{p},(B) \to \mathbf{q}$, $(C) \to \mathbf{s},(D) \to \mathbf{p}$

2. Column II gives approximate values of magnetic fields due to source given in column I

	Column I		Column II
(A)	At surface of neutron star	(p)	$10^{-10} T$
(B)	Near big electromagnet	(q)	$1.5 T$
(C)	At earth surface	(r)	$10^{8} T$
(D)	In interstellar space	(s)	$10^{-4} T$

Show Answer

Answer: $(A) \to \mathbf{r},(B) \to \mathbf{q}$, $(C) \to \mathbf{s},(D) \to \mathbf{p}$

3. Equal currents i flow in two wires along $x$ and $y$ axis as shown. Match the following :

	Column I		Column II
(A)	Magnetic field in first quadrant	(p)	inwards
(B)	Magnetic field in second quadrant	(q)	outwards
(C)	Magnetic field in third quadrant	(r)	may be inwards or outwards
(D)	Magnetic field in fourth quadrant

Show Answer

Answer: $(A) \to \mathbf{r},(B) \to \mathbf{q}$, $(C) \to \mathbf{r},(D) \to \mathbf{p}$

Very Short Answer Questions:

DIRECTIONS : Give answer in one word or one sentence.

1. Why does a compass needle get deflected when brought near a bar magnet?

Show Answer

Answer: A compass needle gets deflected when brought near a bar magnet because magnetic force is exerted by the bar magnet on the compass needle, which is itself a tiny pivoted magnet.

2. A current through a horizontal power line flows in east to west direction. What is the direction of magnetic field at a point directly below it and at a point directly above it ?

Show Answer

Answer: The current is in the east-west direction. Applying the righthand thumb rule, we get that the direction of magnetic field at a point below the wire is from north to south. The direction of magnetic field at a point directly above the wire is from south to north.

3. Why don’t two magnetic lines of force intersect each other?

Show Answer

Answer: Two magnetic lines of force never intersect each other. If the lines intersect, then at the point of intersection there would be two directions (the needle would point towards two directions) for the same magnetic field, which is no possible.

4. Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.

Show Answer

Answer: Direction of magnetic field inside the loop - Perpendicular to the plane of paper inward and Direction of magnetic field outside the loop-Perpendicular to the plane of paper outward.

5. List three sources of magnetic fields.

Show Answer

Answer: (a) Natural and artificial magnets.

(b) Electromagnets

(c) $A$ current carrying conductor produces magnetic field.

6. When is the force experienced by a current-carrying conductor placed in a magnetic field largest?

Show Answer

Answer: The force experienced by a current-carrying conductior placed in a magnetic field is largest when the direction of current is at right-angles to the direction of the magnetic field.

7. What do you conclude from Oersted’s experiment?

Show Answer

Answer: Oersted’s experiment demonstrates the magnetic effect of current. It showed experimentally that the current flowing through a conductor produces a magnetic field.

8. Name the types of electromagnets commonly used.

Show Answer

Answer: (i) Bar type

(ii) Horse shoe type

9. When can an electric charge give rise to a magnetic field?

10. Why is soft iron used as the core of the electromagnet used in electric bell?

11. How will the direction of force be changed, if the current is reversed in the conductor placed in a magnetic field?

12. Describe a set up for plotting the magnetic lines of force in a straight conductor.

13. What is the direction of magnetic field at the centre of a coil carrying current in (i) clockwise (ii) anticlockwise direction?

Show Answer

Answer: (i) Along the axis of the coil inwards

(ii) Along the axis of the coil outwards.

14. Why does a current carrying, freely suspended solenoid rest along a particular direction?

15. What constitutes the field of a magnet?

Show Answer

Answer: Magnetic field lines at any point constitute the field of a magnet indicating that a north pole experiences a force at that point

16. Name the physical quantity whose S.I. unit is $Wb-m^{2}$. Is it a scalar quantity or vector quantity?

Show Answer

Answer: Magnetic field. It is a vector quantity.

17. Name the rule used to find the direction of force on a current carrying conductor.

Show Answer

Answer: Fleming’s left hand rule.

18. What is the greatest disadvantage with a suspended-type moving coil galvanometer?

Show Answer

Answer: It is not portable.

19. If the length of the suspension wire is increased does the sensitivity of a moving coil galvanometer increase or decrease?

Show Answer

Answer: Sensitivity increases.

20. If a mirror of greater radius of curvature replaces the mirror of a moving coil galvanometer, does the sensitivity of the galvanometer, does the sensitivity of the galvanometer increase or decrease?

Show Answer

Answer: It remains unchanged.

21. Why are the pole pieces of a galvanometer made concave?

Show Answer

Answer: The concave pole pieces provide a radial magnetic field, which is strong.

Short Answer Questians :

DIRECTIONS : Give answer in 2-3 sentences.

1. State two ways through which the strength of an electromagnet can be increased.

2. State three factors on which, the magnitude of force on a current carrying conductor placed in a magnetic field, depends. Can this force be zero for some position of the conductor?

3. What do you mean by an electromagnet? With the help of diagrams show the two types of electromagnets. Give two uses of electromagnets.

4. How will you experimentally show that a current-carrying conductor experiences a force when kept in a magnetic field?

5. Under what conditions permanent electromagnet is obtained if a current carrying solenoid is used? Support your answer with the help of a labelled circuit diagram.

Show Answer

Answer: (i) The current through the solenoid should be direct current.

(ii) The rod inside is made of a magnetic material such as steel.

6. $AB$ is a current carrying conductor in the plane of the paper as shown in Figure. What are the directions of magnetic fields produced by it at points $P$ and $Q$ ? Given $r_1>r_2$, where will the strength of the magnetic field be larger?

Show Answer

Answer: Into the plane of paper at $P$ and out of it at $Q$. The strength of the magnetic field is larger at the point located closer i.e. at $Q$.

7. Meena draws magnetic field lines of field close to the axis of a current carrying circular loop. As she moves away from the centre of the circular loop she observes that the lines keep on diverging. How will you explain her observation.

Show Answer

Answer: Strength of the magnetic field falls as distance increases. This is indicated by the decrease in degree of closeness of the lines of field.

8. What does the divergence of magnetic field lines near the ends of a current carrying straight solenoid indicate?

Show Answer

Answer: The divergence, that is, the falling degree of closeness of magnetic field lines indicates the fall in strength of magnetic field near and beyond the ends of the solenoid.

9. List the properties of magnetic lines of force.

Show Answer

Answer: (i) The direction of the magnetic field is indicated by the \to in the line at any point (Tangent).

(ii) The field lines come out of the north pole and get into the south pole (closed loops are formed).

(iii) The strength of magnetic field is indicated by the closeness of the field lines. Closer the lines, more will be the strength and farther the lines, lesser will be the field strength.

(iv) No two field lines will intersect each other if they intersect there will be two different directions for field at the same point which is not impossible.

10. What is the magnetic field that exerts a force of $2.4 \times 10^{-4}$ newtons on a current of 12 amperes in a wire $30 cm$ long set perpendicular to the field?

Show Answer

Answer: The current is perpendicular to the field, $B=\frac{F}{I \ell}$

Therefore, $B=\frac{2.4 \times 10^{-4} N}{(12 A)(0.30 m)}=6.7 \times 10^{-5} N / A . m$

Since $1 mT$ is $10^{-3}$ N/A.m, the answer can be written as 0.067 mT

11. A wire carrying $1.5 amp$ has a length of $20 cm$ in a magnetic field of 40 milliteslas. If the wire is perpendicular to the field, how much force does the field exert on the wire?

Show Answer

Answer: With the wire perpendicular to the field,

$\quad$ F = I/B = (1.5A) (0.20m) (0.040T) = 0.012N

12. What is the direction of the force that a vertical magnetic field, directed upward, will exert on an electron traveling eastward in it.

Show Answer

Answer: Point you left fingers upward, since your are dealing with a negative charge. Rotate your hand until the thumb points east. Your palm will point northward, and that is the direction of the force.

13. What is the shape of magnetic field lines around a circular current carrying conductor?

Show Answer

Answer: Concentric lines around the two ends of the conductor and straight lines at the centre of the conductor.

14. What is Solenoid?

Show Answer

Answer: A cylindrical coil of many tightly wound turns of insulated wires with generally diameter of the coil smaller than its length is called a solenoid.

15. Why is soft iron generally used as the core of the electromagnet?

Show Answer

Answer: (i) Soft iron has less retentivity so it acquires the magnetic properties cnly when the current flows through the coil wound on it and loses the magnetic properties as the current is switched off.

(ii) The soft iron intensifies the magnetic field of the electromagnet because of its high permeability.

16. State four factors on which the strength of electromagnet depends.

Show Answer

Answer: Factors on which the strength of an electromagnet depends are:

(1) Strength of electromagnet is directly proportional to the diameter of coil (Area of cross-section).

(2) Strength of an electromagnet depends upon the nature of the core.

(3) Strength of an electromagnet is directly proportional to the number of turns in coil.

(4) Strength of electromagnet is directly proportional to current.

Long Answer Questians :

DIRECTIONS : Give answer in four to five sentences.

1. What is the nature of magnetic field produced by a straight current carrying conductor? Explain with the help of an experiment.

Show Answer

Answer: The magnetic field produced around a straight current carrying conductor is in the form of concentric circles with the centre lying on the straight conductor.

Take a copper wire AB. Pass it through a cardboard. Connect the wire to a battery through a key. Sprinkle some iron filings on the cardboard. Switch on the key and tap the cardboard gently. You will find that the iron filings arrange themselves in the form of concentric circles. Reverse the direction of current by changing the polarity of the battery. You will find that this time too, the iron filings arrange themselves in concentric circle but in opposite direction. Hence, the magnetic field lines of force around a straight conductor carrying electric current are concentric circles with the conductor at the centre. The direction of magnetic field changes if the direction of current is reversed.

2. Diagram below shows a circuit containing a coil wound over a long and thin hollow carboard tube. Copy the diagram (i) Show the polarity acquired by each face of the solenoid. (ii) Draw the magnetic field lines of force inside the coil and also show their direction. (iii) Mention two methods to increase the strength of the magnetic field inside the coil.

Show Answer

Answer: (i) The polarity acquired by the two ends is as shown below. (A shows North and B shows South polarity

(ii)

(iii) Increase the strength of current, increase the number of turns in the coil, insert soft iron rod in the coil.

3. A 0.4m wire, stretched horizontally, carries an electric current of $15 A$ from east to west, in a magnetic field whose magnetic field intensity is $0.1 N / Am$, directed vertically downwards. What is

(a) the magnitude of the magnetic deflecting force on the wire, and

(b) its direction?

Show Answer

Answer: (a) Given : Length of the wire $(\ell)=0.4 m$

Current $(I)=15 A$

Magnetic induction $(B)=0.1 N / Am$

To calculate : (i) Force $(F)=$ ? (ii) Direction $=$ ?

Formula to be used : $F=BI \ell$

Substituting the given values,

$F=0.1 \times 15 \times 0.4=0.6 N$

(b) By Fleming’s left hand rule, forefinger (magnetic field) points vertically downwards, the middle finger (current) points west and the thumb (force) points towards the south.

4. A straight conductor passes vertically through a cardboard sprinkled with iron filings. Show the setting of the iron filings when a weak current is passed in the downward direction. What changes occur if,

(i) the strength of the current is increased.

(ii) the single conductor is replaced by several parallel conductors with current flowing in the same. direction.

Show Answer

Answer: Figure shows the setting of the iron filings.

(i) The shape of distribution of iron filings remains unchanged but they get arranged upto a larger distance from the conductor when the strength of current is increased. This is because on increasing the strength of current, the strength of the magnetic field is increased and it is effective upto a larger distance

(ii) Magnetic field strength is increased so the iron filings get arranged upto a larger distance.

5. The diagram shows a current carrying coil passing through a sheet of stiff cardboard. Draw three lines of magnetic field on the cardboard.

Show Answer

Answer: Figure represents the magnetic lines of force due to current carrying coil.

The magnitude of magnetic field at the centre depends on (i) the strength of current in the coil and (ii) the radius of coil.

State two factors on which the magnitude of magnetic field at the centre of coil, depends.

6. Draw a labelled diagram to make an electromagnet from a soft iron bar. Mark the polarity at its ends. What precaution would you observe?

Show Answer

Answer: The labelled diagram is shown in figure.

Precaution : The source of current must be the d.c. source.

2 EXERCISE

Multiple Choice Questians :

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

1. Choose the correct option(s).

The magnetic field inside a long straight solenoid-carrying current

(a) is zero.

(b) decreases as we move towards its end.

(c) increases as we move towards its end.

(d) is the same at all points.

Show Answer

Answer: (d)

2. A positively-charged particle (alpha-particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is

(a) towards south

(b) towards east

(c) downward

(d) upward

Show Answer

Answer: (d)

3. Which of the following correctly describes the magnetic field near a long straight wire?

(a) The field consists of straight lines perpendicular to the wire.

(b) The field consists of straight lines parallel to the wire

(c) The field consists of radial lines originating from the wire.

(d) The field consists of concentric circles centred on the wire.

Show Answer

Answer: (d)

4. A magnet is placed vertically on a paper. Then the number of neutral points obtained on the paper is

(a) zero

(b) one

(c) two

(d) three

Show Answer

Answer: (b)

5. A small magnet is placed perpendicular to a uniform magnet field. The forces acting on the magnetic will result in

(a) Rotational motion

(b) Translatory motion

(c) No motion at all

(d) Translational and rotational motion both

Show Answer

Answer: (a)

6. A charge of $1.6 \times 10^{-19}$ coulomb enters in the magnetic field of 2 weber $/ m^{2}$ normally with a velocity of $10^{5} m / sec$. The force on the charge will be

(a) $3.2 \times 10^{-14} N$

(c) $1.6 \times 10^{-14} N$

(b) $3.2 \times 10^{-19} N$

(d) zero

Show Answer

Answer: (a)

7. The magnetic field at a point due to a current carrying conductor is directly proportional to the

(a) current flowing through the conductor

(b) Distance from the conductor

(c) Voltage across the conductor

(d) Resistance of the conductor

Show Answer

Answer: (a)

8. Which one of the following substances is the magnetic substances?

(a) Mercury

(b) Iron

(c) Gold

(d) Silver

Show Answer

Answer: (b)

9. Magnetic lines do not intersect one-another because

(a) they are at a distance

(b) they are in the same direction

(c) they are parallel to another

(d) at the point intersection there will be two direction of the magnetic force which is impossible

Show Answer

Answer: (d)

10. Instrument can be shielded from outside magnetic effects by surrounding them with

(a) Rubber shield

(b) Glass shield

(c) Iron shield

(d) Brass shield

Show Answer

Answer: (c)

11. The vertical plane which passes through the magnetic axis of a freely suspended magnetic at rest is called

(a) Magnetic meridian

(b) Geographical meridian

(c) North meridian

(d) South meridian

Show Answer

Answer: (a)

12. By removing the inducing magnet, the induced magnetism is

(a) Finished after some time

(b) Finished just after

(c) Not finished for a long time

(d) Not charged

Show Answer

Answer: (b)

13. The similar magnets of steel are than the magnets of soft iron

(a) stronger

(b) of equal strength

(c) weaker

(d) none of the above

Show Answer

Answer: (c)

14. A bar of soft iron is placed flat on the table. A bar magnet is taken and its south pole is placed on one end of the bar of soft iron. The magnet is held almost vertically. The bar is stroked from one end to the other with magnet. On the other end of the bar, magnet is lifted and again placed on the first end and the bar is again stroked. The end of the bar where the magnet is lifted will be

(a) south pole

(b) no pole

(c) south and north both type

(d) north pole

Show Answer

Answer: (d)

15. The angle of decilination at a place is the angle

(a) between the vertical plane and the geographical meridian

(b) between the vertical plane and the magnetic meridian

(c) between the geographical meridian and the magnetic meridian

(d) between the geographical meridian and horizontal plane

Show Answer

Answer: (c)

16. The magnetism in a magnet is mainly due to

(a) The orbital motion of the electrons

(b) The spin motion of the electrons

(c) The nuclear charge

(d) None of the above

Show Answer

Answer: (b)

17. When the north pole of a strong magnet is brought near to the north pole of a weak magnet then they will

(a) Attract each other

(b) repel each other

(c) first attract and then repel

(d) first repel and then attract

Show Answer

Answer: (a)

18. A magnet can be demagnetised by

(a) Hammering the magnet

(b) Putting it in the water

(c) Cooling it

(d) Putting in contact with iron

Show Answer

Answer: (a)

19. The effective length of the magnet is

(a) the complete length of the magnet

(b) the distance between the two poles of the magnet

(c) the half of the length of the magnet

(d) the square of the length of the magnet

Show Answer

Answer: (b)

20. When the bars of bismuth are placed between the magnetic poles they set their length

(a) perpendicular to the lines of force

(b) along the lines of force

(c) neither perpendicular nor along the lines of force

(d) In any direction

Show Answer

Answer: (a)

21. Two bars of soft iron exactly same are given. One of them is a magnet. Without using any thing more, how would you find which is a magnet

(a) By bringing two bars near and noting which one is attracting. The attracting one is a magnet

(b) By bringing two bars near and noting which one is repelling. One which repells is an ordinary iron.

(c) By rubbing one bar with the other and noting which becomes magnet. The bar which is magnetised is an ordinary iron

(d) One bar is placed flat horizontal on the table and the other bar is held vertical with its one end on the middle of first bar. If there is attraction between the two, the vertical bar is magnet otherwise ordinary iron.

Show Answer

Answer: (d)

22. Magnetic storms are due to

(a) the rotation of the earth

(b) the revolution of the earth

(c) the rainy season

(d) the appearance off sun spots

Show Answer

Answer: (d)

23. Which of the following processes will not produce new magnetic poles?

(a) cutting a bar magnet in half

(b) turning on a current in a solenoid

(c) running a current through a straight wire

(d) placing an iron rod in contact with a magnet

Show Answer

Answer: (c)

24. Magnetic field lines start

(a) on $N$-poles

(b) on S-poles

(c) on current-carrying wires

(d) Nowhere

Show Answer

Answer: (d)

25. The conductivity of a magnetic substance for the lines of force with respect to air is called the

(a) Magnetic induction

(b) magnetic permeability

(c) magnetic flux density

(d) intensity of magnetisation

Show Answer

Answer: (b)

26. Magnetic field lines form circles in the space

(a) near a permanent magnet

(b) around a current-carrying wire

(c) inside a solenoid

(d) inside a current-carrying loop

Show Answer

Answer: (b)

27. The angle of inclination of the axis of the magnetic needle to the horizontal plane when suspended freely and is at rest is known as

(a) angle of inclination

(b) angle of variation

(c) angle of declination

(d) none of the above

Show Answer

Answer: (a)

28. A tesla is equivalent to $a$

(a) newton per coulomb

(b) newton per ampere-meter

(c) ampere per newton

(d) newton per ampere-second

Show Answer

Answer: (b)

29. A vertical wire carries a current upward. The magnetic field north of the wire will be directed

(a) upward

(c) westward

(b) eastward

(d) northward

Show Answer

Answer: (c)

30. The magnetic flux is expressed in

(a) dynes

(b) Oerested

(c) Gauss

(d) Weber

Show Answer

Answer: (d)

31. The magnetic lines of force, inside a current carrying solenoid, are

(a) along the axis and are parallel to each other

(b) perpendicular to the axis and equidistance from each other

(c) circular and they do not intersect each other

(d) circular at the ends but they are parallel to the axis inside the solenoid.

Show Answer

Answer: (a)

32. In a moving coil galvanometer, the magnetic pole pieces are made cylindrical and a soft iron core is placed at the centre of the coil to make the magnetic field

(a) strong

(b) strong and radial

(c) uniform

(d) strong and uniform

Show Answer

Answer: (b)

33. Which of the following is not true

(a) Induction precedes attraction

(b) We cannot isolate a single pole

(c) We can magnetise an iron ring

(d) A permanent magnet retains its magnetism even when heated on a flame.

Show Answer

Answer: (d)

34. Wrist watches are made antimagnetic by shielding their machinery with

(a) plastic sheets

(b) a metal of high conductivity

(c) a magnetic substance of low permeability

(d) a magnetic substance of high permeability

Show Answer

Answer: (d)

35. Which of the following statement is not correct?

(a) The dip angle is the angle between the horizontal and earth’s total magnetic field.

(b) Neutral points are obtained where the earth’s magnetic field is perpendicular to that due to a magnet

(c) A magnetic field is a region in which a magnetic force can be detected.

(d) Magnetic fields are vector quantities

Show Answer

Answer: (b)

36. When a bar magnet is broken into two pieces?

(a) we will have a single pole on each piece

(b) each piece will have two like poles

(c) each piece will have two unlike poles

(d) each piece will be lose magnetism

Show Answer

Answer: (c)

37. The permanent magnets are kept with soft iron pieces at ends as keepers

(a) to magnetise the soft iron pieces

(b) to increase the strength of the magnets

(c) to avoid self demagnetisation

(d) for physical safety of the magnets

Show Answer

Answer: (c)

38. A small magnet is placed perpendicular to a constant magnetic field. The forces acting on the magnet will result in

(a) rotation

(b) translation

(c) no motion at all

(d) rotation as well as translation

Show Answer

Answer: (d)

39. The vertical component of the earth’s magnetic field is

(a) zero at the magnetic pole

(b) zero at the geographic pole

(c) same everywhere

(d) zero at the magnetic equator

Show Answer

Answer: (d)

40. A small piece of a substance gas repelled when it is brought near a powerful magnet. The substance can be

(a) dimagnetic

(b) paramagnetic

(c) ferromagnetic

(d) non-magnetic

Show Answer

Answer: (a)

41. Which of the following statement is not correct about two parallel conductors carrying equal currents in the same direction?

(a) Each of the conductors will experience a force

(b) The two conductors will repel each other.

(c) There are concentric lines of force around each conductor

(d) Each of the conductors will move if not prevented from doing so

Show Answer

Answer: (b)

42. Which of the following determines the direction of magnetic field due to a current carrying conductor?

(a) Faraday’s laws of electromagnetic induction

(b) Fleming’s left-hand rule

(c) Lenz’s rule

(d) Maxwell’s cork screw rule

Show Answer

Answer: (d)

43. Along the direction of current carrying wire, the value of magnetic field is?

(a) Zero

(b) Infinity

(c) Depends on the length of the wire

(d) Uncertain

Show Answer

Answer: (a)

44. To obtain maximum intensity of magnetic field at a point the angle between position vector of point and small elements of length of the conductor is

(a) 0

(b) $\pi / 4$

(c) $\pi / 2$

(d) $\pi$

Show Answer

Answer: (c)

45. The value of magnetic field due to a small element of current carrying conductor at a distance $r$ and lying on the plane perpendicular to the element of conductor is

(a) zero

(b) maximum

(c) inversely proportional to the current

(d) none of the above

Show Answer

Answer: (b)

46. The value of intensity of magnetic field at a point due to a current carrying conductor depends

(a) Only on the value of current

(b) Only on a small part of length of conductor

(c) On angle between the line joining the given point to the mid point of small length and the distance between the small length of the point

(d) On all and the above

Show Answer

Answer: (d)

47. An electric current $i$ is flowing in a circular coil of radius a. At what distance from the center of the axis of the coil will the magnetic field be $1 / 8^{\text{th }}$ of its value at the centre?

(a) $3 a$

(b) $\sqrt{3}$ a

(c) $a / 3$

(d) $a / \sqrt{3}$

Show Answer

Answer: (b)

48. In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite direction. The magnetic field is zero-

(a) outside the cable

(b) inside the inner conductor

(c) inside the outer conductor

(d) in between the two conductors

Show Answer

Answer: (a)

49. When the number of turns in a toroidal coil is doubled, the value of magnetic flux density will become

(a) four times

(b) eight times

(c) half

(d) double

Show Answer

Answer: (d)

50. By a current carrying toroid, whose area of cross-section is fixed, the magnetic induction produced will be -

(a) maximum at inner end

(b) maximum at outer end

(c) maximum at center of area of cross-section

(d) equal at whole area of cross-section

Show Answer

Answer: (d)

51. When an electron beam is moving in a magnetic field, then the work done is equal to the

(a) charge of electron

(b) magnetic field

(c) product of electronic charge and the magnetic field

(d) zero

Show Answer

Answer: (d)

52. Which of the following rays are not deflected by a magnetic field -

(a) $\alpha$-rays

(b) $\beta$-rays

(c) $\gamma$-rays

(d) positive rays

Show Answer

Answer: (c)

53. Two parallel beams of electron moving in the same direction will

(a) repel each other

(b) attract each other

(c) not interact with each other

(d) annihilate each other

Show Answer

Answer: (b)

54. An inverse square law of distance is obeyed by-

(a) the force per unit length between two long straight parallel current-carrying conductors in vacuum.

(b) the electric field intensity outside an isolated charged sphere

(c) the magnetic flux density outside a long straight current-carrying wire

(d) the electrostatic potential

Show Answer

Answer: (b)

55. A current carrying loop lying in a magnetic field behaves like a.

(a) A magnetic dipole

(b) magnetic pole

(c) magnetic material

(d) non-magnetic material

Show Answer

Answer: (a)

More Than One Carrect :

DIRECTIONS : This section contains multiple choice questions. Each question has 4 choices $(a),(b),(c)$ and (d) out of which ONE OR MORE may be correct.

1. Which of the following is/are valid for a current carrying circular coil?

(a) To induce magnetic induction of constant magnitude (at the centre of the coil), current required in it is directly proportional to its radius

(b) Induction of inducted magnetic field at the centre is inversely proportional to the radius of the coil

(c) Induction of the inducted magnetic field at the centre is directly proportional to the current flowing through the directly proportional to the current flowing through the coil

(d) None of the above

Show Answer

Answer: (a,b,c)

2. A current flows along the length of a long thin cylindrical shell:

(a) magnetic field at all the points lying inside the shell is zero

(b) magnetic field at any point outside the shell varies inversely with distance from the surface of the shell

(c) magnetic field strength is maximum just outside the shell

(d) none of the above

Show Answer

Answer: (a,c)

3. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If $E$ and $B$ represent the electric and magnetic fields repectively, this region of space may have:

(a) $E=0, B=0$

(b) $E=0, B \neq 0$

(c) $E \neq 0, B=0$.

(d) $E \neq 0, B \neq 0$

Show Answer

Answer: (a,d)

4. Figure shows a bar magnet and a long straight wire $W$ carrying current into the plane of the paper. Point $P$ is the point of inersection of axis of the magnet and the line of the shortest distance between the magnet and the wire.

Which of the following statements is/are correct?

(a) If $P$ is the midpoint of the magnet, the magnet experiences no torque

(b) If $P$ is on the left of the midpoint of the magnet, the magnet experiences a leftward force as well as a torque

(c) If $P$ is the midpoint of the megnet, the magnet experience a force along the line of the shortest distance

(d) None of the above

Show Answer

Answer: (a,b,c)

5. A straight conductor carries a current. Assume that all free electrons in the conductor move with the same drift velocity $v$. $A$ and $B$ are two observers on a straight line $X Y$ parallel to the conductor. $A$ is stationary. B moves along $X Y$ with a velocity $v$ in the direction of the free electrons.

(a) $A$ and $B$ observe the same magnetic field

(b) $A$ observes a magnetic field, $B$ does not

(c) $A$ and $B$ observe magnetic fields of the same magnitude but opposite directions

(d) $A$ and $B$ do not observe any electric field

Show Answer

Answer: (a,d)

6. A charged particle enters into a space and continues to move undeflected. Then in that space:

(a) a uniform horizontal electric field and a vertical magnetic field may be present

(b) a vertical electric field alone may be present

(c) uniform electric and magnetic fields, both directed vertically downwards, may be present

(d) a uniform horizontal magnetic field alone may be

Show Answer

Answer: (b,c,d)

7. A long straight conductor, carrying a current $I$, is bent to form an almost complete circular loop of radius $r$ on it. The magnetic field at the centre of the loop:

(a) is directed into the paper

(b) is directed out of the paper

(c) has magnitude $\frac{\mu_0 I}{2 r}(1-\frac{1}{\pi})$

(d) has magnitude $\frac{\mu_0 I}{2 r}(1+\frac{1}{\pi})$

Show Answer

Answer: (a,c)

8. L is a circular ring made of a uniform wire. Current enters and leaves the ring through straight conductors which, if produced, would have passed through the centre $C$ of the ring. The magnetic field at $C$ :

(a) due to the straight conductors is zero

(b) due to the loop is zero

(c) due to the loop is proportional to $\theta$

(d) due to the loop is proportional to $(\pi-\theta)$

Show Answer

Answer: (a,b)

9. A flat circular coil carrying a current has a magnetic moment $\mu$ :

(a) $\mu$ has only magnitude; it does not have direction

(b) the direction of $\mu$ is along the normal to the plane of the coil

(c) the direction of $\mu$ depends on the direction of current flow

(d) the direction of $\mu$ does not change if the current in the coil is reversed

Show Answer

Answer: (b,c)

10. A current carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring:

(a) there is no net force on the ring

(b) the ring will tend to expand

(c) the ring will tend to contract

(d) either (b) or (c) depending on the directions of the current in the ring and the magnetic field

Show Answer

Answer: (a,d)

11. A solenoid is connected to a source of constant e.m.f. for a long time. A soft iron piece is inserted into it. Then:

(a) self-inductance of the solenoid gets increased

(b) flux linked with the solenoid increases; hence, steady state current gets decreased

(c) energy stored in the solenoid gets increased

(d) magnetic moment of the solenoid gets increased

Show Answer

Answer: (a,c,d)

12. In the loops shown in figure all curved sections are either semicircles or quarter circles. All the loops carry the same current. The magnetic fields at the centres have magnitudes $B_1, B_2, B_3$ and $B_4$ :

(a) $B_4$ is maximum

(c) $B_4>B_1>B_2>B_3$

(b) $B_3$ is minimum

(d) $B_1>B_4>B_3>B_2$

Show Answer

Answer: (a,b,c)

Fill in the Passage :

DIRECTIONS : Fill in the blanks in the following passage(s) from the words given inside the box.

I.

Oersted was the first to put forth the direct relation between ………1…….. and magnetism. He conducted several experiments to determine the ………2…….. of a current ………3…….. wire. The following describes the Oersted experiment conducted to establish that a current carrying wire acts as a ………4…….. .

A long straight wire is connected to an external battery and an electric current is passed through it. When a magnetic needle is placed below the wire such that the wire is parallel to the axis of the magnetic needle and the current flows in the south to north direction, a deflection in the needle is observed. It is observed that the north pole of the needle is deflected westwards and as the ………5…….. is increased, the deflection, ………6…….. till the north pole of the needle turns towards exact west. It is also observed that if instead of placing the magnetic needle below the wire, if it was placed above the wire, the north pole of the magnetic needle is deflected . ………7…….. .

Show Answer

Answer: (1) electricity $\quad$ (2) magnetic effect (3) carrying $\quad$ (4) magnet (5) magnitude of current $\quad$ (6) increases $\quad$ (7) eastwards

II.

A magnet ………1…….. small pieces of iron, nickel, cobalt, etc. The magnetic force of attraction is maximum is small regions near the ends of the magnet. These are called the ………2…….. of the magnet.

A freely suspended magnet always points in ………3…….. direction. The end pointing towards north is called north seeking pole or north pole. The other end which points towards south is called south seeking pole or south pole.

All nagnets attract ………4…….. poles of other magnets and also a few substances like iron, nickel and cobalt. Thus if an unknown substance is attracted by a magnet, we cannot be sure whether it is a magnet. But like poles of magnet ………5…….. each other. Hence if there is repulsion between an unknown substance and a magnet, we can be sure that the unknown substance is a magnet. The substances and a magnet, we can be sure that the unknown substance is a magnet. The substances which are attracted by magnets are called ………6…….. substances.

Show Answer

Answer: (1) attracts $\quad$ (2) poles $\quad$ (3) north-south $\quad$ (4) opposite $\quad$ (5) repel $\quad$ (6) magnetic

III.

The magnetic field produced around a straight current carrying conductor is in the form of ………1…….. with the ………2…….. lying on the straight conductor.

Take a copper wire AB. Pass it through a cardboard. Connect the wire to a battery through a key. Sprinkle some iron filings on the cardboard. Switch on the key and tap the cardboard gently. You will find that the iron filings arrange themselves in the form of concentric circles. Reverse the direction of current by changing the ………3…….. of the battery. You will find that this time too, the iron filings arrange themselves in concentric circle but in ………4…….. direction.

Hence, the magnetic field lines of force around a ………5…….. carrying electric current are concentric circles with the conductor at the centre. The direction of magnetic field changes if the direction of current is ………6……..

Show Answer

Answer: (1) Concentric circles $\quad$ (2) centre $\quad$ (3) polarity $\quad$ (4) opposite $\quad$ (5) straight conductor $\quad$ (6) reversed

Passage Based Questions :

Passage-I

An electron with kinetic energy $K_e$ travels in a circular path that is perpendicular to a uniform magnetic field, the electron’s motion subject only to the force due to the field.

1. The magnetic dipole moment of the electron due to its orbital motion has magnitude :

(a) $K_e B$

(b) $\frac{K_e B}{2}$

(c) $K_e / B$

(d) $K_e / 2 B$

Show Answer

Answer: (c)

2. The magnetic dipole moment of a positive ion with kinetic energy $K$ under the same circumstances :

(a) $K_i B$

(b) $K_i / B$

(c) $K_i / 2 B$

(d) Zero

Show Answer

Answer: (b)

3. An ionized gas consists of $5.3 \times 10^{21}$ electrons $/ m^{3}$ and the same number density of ions. Take the average electron kinetic energy to be $6.2 \times 10^{-20} J$ and average ion kinetic energy to be $7.6 \times 10^{-21} J$. The magnetisation of the gas when it is in a magnetic field of $1.2 T$ is

(a) $150 A / m$

(b) $215 A / m$

(c) $295 A / m$

(d) $310 A / m$

Show Answer

Answer: (d)

Passage-II

Take a tightly-wound solenoid of radius $a$ and length $\ell$. the number of turns per unit length in it is $n$. It carries current $i$. Consider a small element of length $d x$ of the solenoid at a distance $x$ from one end. This contains ndx turns and can be assumed as a current carrying loop. The magnetic field due to whole solenoid will be the sum of magnetic field due to such elements.

1. The magnetic field due to this solenoid at the centre of its axis is

(a) $\frac{\mu_0 i}{2 a}$

(b) $\frac{\mu_0 n i}{2 a}$

(c) $\frac{\mu_0 n i}{\sqrt{1+\frac{a^{2}}{\ell^{2}}}}$

(d) $\frac{\mu_0 n i}{\sqrt{1+\frac{4 a^{2}}{\ell^{2}}}}$

Show Answer

Answer: (d)

2. The magnetic field due to the solenoid at the centre of its axis in the situation $a \gg \ell$ is :

(a) $\frac{\mu_0 n i}{2}$

(b) $\mu_0 n i$

(c) $\frac{\mu_0 n i \ell}{2 a}$

(d) $\frac{\mu_0 n i \ell}{a}$

Show Answer

Answer: (c)

3. The magnetic field due to the solenoid at the centre of its axis in the situation $a \ll \ell$ is :

(a) zero

(b) $\frac{\mu_0 n i}{2}$

(c) $\mu_0 n i$

(d) infinite

Show Answer

Answer: (c)

Passage-III

In a television tube, each of the electrons in the beam has a kinetic energy of $12.0 keV$. The tube is oriented so that the electrons move horizontally from geomagnetic south to geomagnetic north. The vertical component of earth’s magnetic field points down and has a magnitude of $55.0 \mu T$.

1. The direction in which beam deflects :

(a) east

(b) west

(c) north-eas

(d) south-west

Show Answer

Answer: (a)

2. The acceleration of any electron due to the magnetic field is

(a) $3.14 \times 10^{14} m / s^{2}$

(b) $4.28 \times 10^{14} m / s^{2}$

(c) $5.56 \times 10^{12} m / s^{2}$

(d) $6.28 \times 10^{14} m / s^{2}$

Show Answer

Answer: (d)

3. The transverse deflection of the beam after travelling 20.0 $cm$ through the television tube :

(a) $1.96 mm$

(b) $2.98 mm$

(c) $4.24 mm$

(d) none of these

Show Answer

Answer: (b)

Assertion & Reason :

DIRECTIONS : Each of these questions contains an Assertion followed by reason. Read them carefully and answer the question on the basis of following options. You have to select the one that best describes the two statements.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(c) If Assertion is correct but Reason is incorrect .

(d) If Assertion is incorrect but Reason is correct .

1. Assertion : A direction current flows through a metallic rod, produced magnetic field only outside the rod.

Reason : There is no flow of charge carriers inside the rod.

Show Answer

Answer: (e) In the case of metallic rod, the charge carries flow through whole of the cross section. Therefore, the magnetic field exists both inside as well as outside. However magnetic field inside the rod will go no decreasing as we go towards the axis.

2. Assertion : A proton moves horizontally towards a vertical long conductor having an upward electric current. It will deflect vertically downward.

Reason : Seeing the proton and the conductor from the side of the proton, the magnetic field at the site of the proton will be towards right. Hence the force $\overrightarrow{F}=q \overrightarrow{v} \times \overrightarrow{B}$ will deflect the proton vertically downward.

Show Answer

Answer: (a)

3. Assertion : Force experienced by moving charge will be maximum if direction of velocity of charge is perpendicular to applied magnetic field.

Reason : Force on moving charge is independent of direction of applied magnetic field.

Show Answer

Answer: (c) From equation F = quB sin $\theta$. Force on moving charge will be maximum if direction of velocity of charge is perpendicular to direction of magnetic field (when $\theta = 90^ \circ $)

4. Assertion : A neutral body may experience a net nonzero magnetic force.

Reason : The net charge on a current carrying wire is zero, but it can experience a force in a magnetic field.

Show Answer

Answer: (a)

5. Assertion : There is no change in the energy of a charged particle moving in a magnetic field although a magnetic force is acting on it.

Reason : Work done by centripetal force is always zero.

Show Answer

Answer: (a) Magnetic force is always perpendicular to the direction of motion of charged particle, i.e., work done on the charge particle moving on a circular path in magnetic field is zero.

6. Assertion : When two long parallel wires, hanging freely are connected in series to a battery, they come closer to each other.

Reason : Wires carrying current in opposite direction repel each other.

Show Answer

Answer: (b) The wires are parallel to each other but the direction of current in it is in same direction so they attract each other. If the current in the wire is in opposite direction then wires repel each other. When the currents are in opposite directions, the magnetic forces are reversed and the wires repels each other

7. Assertion : A solenoid tends to expand, when a current passes through it.

Reason : Two straight parallel metallic wires carrying current in same direction attract each other.

Show Answer

Answer: (d) When current flows through a solenoid, the currents in the various turns of the solenoid are parallel and in the same direction. Since the currents flowing through parallel wires in the same direction lead to force of attraction between them, the turns of the solenoid will also attract each other and as a result the solenoid tends to contract.

8. Assertion : In a conductor, free electrons keep on moving but no magnetic force acts on a conductor in a magnetic field.

Reason : Force on free electrons due to magnetic field always acts perpendicular to its direction of motion.

Show Answer

Answer: (c) In a conductor, the average velocity of electrons is zero. Hence no current flows through the conductor. Hence, no force acts on this conductor.

9. Assertion : A small coil carrying current, in equilibrium, is perpendicular to the direction of the uniform magnetic field.

Reason : Torque is maximum when plane of coil and direction of the magnetic field are parallel to each other.

Show Answer

Answer: (b) The torque acting on a coil is given by,

$ \tau=NLAB \sin \theta $

where $\theta$ is the angle between the plane of the coil and the direction of magnetic field. When $\theta=90^{\circ}$, then $\tau=$ 0 . The coil tries to orient itself in this position. Thus in equilibrium, the coil acquires a position, such that its plane makes an angle $90^{\circ}$ with the direction of magnetic field.

10. Assertion : Basic difference between an electric line and magnetic line of force is that former is discontinuous and the later is continuous or endless.

Reason : No electric lines of force exist inside a charged body but magnetic lines do exist inside a magnet.

Show Answer

Answer: (a) In case of the electric field of an electric dipole, the electric lines of force originate from positive charge and, end at negative charge. Since, isolated magnetic lines are closed continuous loops extending throughout the body of magnet, hence they from endless curves.

11. Assertion : No net force acts on a rectangular coil carrying a steady current when suspended freely in a uniform magnetic field.

Reason : Force on coil in magnetic field is always non-zero.

Show Answer

Answer: (c) Force acting on each pair of the opposite sides of the coil are eq

Multiple Matching Quesions :

DIRECTIONS : Following question has four statements $(A, B$, $C$ and $D$) given in Column I and four statements ($p, q, r$ and $s$) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. Match the entries in column I with entries in column II.

1. In column I, the positions of small current carrying loops have been shown and in column II information related to force experienced by coil is given. Match the entries of column I with the entries of column II. (Assume solenoid radius to be small as compared to its length)

	Column-I		Column-II
A.		p.	Attractive
		q.	Repulsive
		r.	Zero
		s.	Initially zero, then starts Increasing

Show Answer

Answer: $A \to r ; B \to p ; C \to p ; D \to s$

The force experienced by a coil in a magnetic field is given by $F=P_m \frac{\partial B}{\partial r}$, where $\frac{\partial B}{\partial r}$ is the increment of B along magnetic dipole moment of contour. You can write the expression for magnetic field due to solenoid at a general point and then differentiate it. From this information, you can have the results. Whether the force is attractive or repulsive can also be found by using the concept of nature of poles induced on the solenoid and coil.

The force comes out to be zero at centre and as we approach it from some outside point, its value increases.

For D: The coil first rotates to allign itself in such a manner so as to link the maximum flux and then the case would be same as that of $C$.

2. In magnetic field, for a charged particle, match the entries of column I with the entries of column II.

	Column I		Column II
A.	Acceleration	p.	may be zero
B.	Velocity	q.	is zero
C.	Speed	r.	may be constant
D.	Kinetic energy	s.	is constant

Show Answer

Answer: $A \to p ; B \to r ; C \to s ; D \to s$

Work done by magnetic force is zero. From work-energy theorem, its speed or kinetic-energy is constant.

3. A circular current carrying loop is placed in $x-y$ plane as shown in figure. A uniform magnetic field $\vec{B}=B_0 \hat{k}$ is present in the region. Match the entries of column I with the entries of column II.

	Column I		Column II
A.	Magnetic moment of the loop is	p.	Zero
B.	Torque on the loop is	q.	maximum
C.	Potential energy of the loop is	r.	along $+ve z$ - axis
D.	Equilibrium of the loop is	s.	stable
		t.	None

Show Answer

Answer: $A \to r ; B \to p ; C \to t ; D \to p$

$\theta$, the angle between $\vec{M}$ and $\vec{B}$ is zero, as both the vectors are along positive $Z$ - direction.

HOTS Subjective Questians :

DIRECTIONS : Answer the following questions.

1. Consider a circular loop of wire lying in the plane of the table. Let the current pass through the loop clockwise. Apply the right-hand rule to find out the direction of the magnetic field inside and outside the loop.

Show Answer

Answer: Direction of magnetic field inside the loop is perpendicular to the plane of paper inward and direction of magnetic field outside the loop is perpendicular to the plane of paper outward.

2. When is the force experienced by a current-carrying conductor placed in a magnetic field largest?

Show Answer

Answer: The force experienced by a current-carrying conductor placed in a magnetic field is largest when the direction of current is at right-angles to the direction of the magnetic field.

3. Two charged particles are projected into a magnetic field that is perpendicular to their velocities. If the charges are deflected in opposite directions, what does this tell you about the particles?

Show Answer

Answer: Since the charged particles are deflected in opposite directions in the same magnetic field, we can say that they have opposite charges, may be electron and proton, for example.

4. A current through a horizontal power line flows in east to west direction. What is the direction of magnetic field at a point directly below it and at a point directly above it ?

Show Answer

Answer: The current is in the east-west direction. Applying the righthand thumb rule, we get that the direction of magnetic field at a point below the wire is from north to south. The direction of magnetic field at a point directly above the wire is from south to north.

5. A magnetic field that varies in magnitude from point to point but has a constant direction (east to west) is set up in a chamber. A charged particle enters the chamber and travels undeflected along a straight path with constant speed. What can you say about the initial velocity of the particle?

6. An electron travelling west to east enters a chamber having a uniform electrostatic field in north to south direction Specify the direction in which a uniform magnetic field shoulu be set up to prevent the electron from deflecting from its straight line path.

7. An electron in passing through a field but no forces acting on it. Under what condition it is possible, that the motion of the electron will be in the (i) electric field (ii) magnetic field?

Show Answer

Answer: (i) In electric field, there is always a force on the moving electron opposite to the direction of field. Thus the force will be zero only if field is zero.

(ii) In magnetic field, the force acting on a moving electron is $F=q \vee B \sin \theta$, it is zero if $\theta=0^{\circ}$ or $180^{\circ}$ i.e., the electron is moving parallel to the direction of magnetic field.

8. The energy of a charged particle moving in a uniform magnetic field does not change. Explain

Show Answer

Answer: When a charged particle is moving in a uniform magnetic field it experiences a force in a direction, perpendicular to its direction of motion. Due to which the speed of the charged particle remains unchanged and work done on it is zero, hence its kinetic energy remains same.

9. A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape?

Show Answer

Answer: It assumes circular shape with its plane normal to the field to maximize flux, since for a given perimeter, a circle encloses greater area than any other shape.

10. An electron does not suffer any deflection while passing through a region. Are you sure that there is no magnetic field? Is the reverse definite?

Show Answer

Answer: If electron passing through a certain region does not suffer any deflection, then we are not sure that there is no magnetic field in that region. This is due to that electron suffers no force when it moves parallel or antiparallel to magnetic field. Thus the magnetic field may exist parallel or antiparallel to the direction of motion of electron.

11. Why is pure iron not used for making permanent magnets? Name one materials used for making permanent magnets. Describe how permanent magnets are made electrically. State two examples of electrical instruments made by using permanent magnets.

Show Answer

Answer: Pure iron is not used for making permanent magnets because it cannot retain their magnetism for long time and used only for electromagnet since alloys of iron and steel have strongly magnetised and have a capacity to hold it for a longer time period, they are used for permanent magnets. Material used for permanent magnet-ALNICO. Formation of a permanent magnet electrically. Permanent magnets can be formed by placing a hard steel rod in the strong uniform magnetic field produced by the solenoid. Steels have the quality to retain its magnetism after switch off the solenoid current.

Permanent magnets are used in

(i) Galvanometer

(ii) Ammeter.

12. The flow of a current in a circular loop of wire creates a magnetic field at its centre. How may existence of this field be detected? State the rule which helps to predict the direction of this magnetic field.

Show Answer

Answer: To detect the presence of the magnetic field created by the current in a circular loop at the centre, one can draw the magnetic field lines with the help of compass needle. The magnetic field lines appear as a straight line at the centre and other lines appear in the same direction without the loop. Direction of field is given by the right hand thumb’s rule. Right hand thumb rule. Hold the wire in your right hand with your extended thumb pointing in the direction of current. Your folded fingers will indicate the direction of magnetic field around the wire.

13. How does the strength of the magnetic field at the centre of a circular coil of wire depend on :

(i) the radius of the coil

(ii) the number of turns of wire in the coil

(iii) the strength of current flowing in the coil ?

Show Answer

Answer: The strength of the magnetic field produced at the centre of a circular coil of radius $r$, having $N$ turns and carrying a

$ B=\frac{\mu_0 N I}{2 r} \text{ testla } $

Thus, the strength of a magnetic field in the coil is

(i) inversely proportional to the radius of the loop.

$ (B \propto \frac{1}{r}) $

(ii) directly proportional to the number of turns. $(B \propto N)$

(iii) directly proportional to the current passing through it.

14. How will the magnetic field intensity at the centre of a circular coil carrying current change, if the current through the coil is doubled and the radius is halved?

Show Answer

Answer: Four times