Thinking Process

Toroid is a hollow circular ring on which a large number of turns of a wire are closely wound. Thus, in such a case magnetic field is only confined inside the body of toroid.

Answer

(c) In case of toroid, the magnetic field is only confined inside the body of toroid in the form of concentric magnetic lines of force and there is no magnetic field outside the body of toroid. This is because the loop encloses no current. Thus, the magnetic moment of toroid is zero.

In general, if we take $r$ as a large distance outside the toroid, then $m \propto \frac{1}{r^{3}}$. But this case is not possible here.

Answer

(a) For the earth’s magnetism, the magnetic field lines of the earth resemble that of a hypothetical magnetic dipole located at the centre of the earth.

The axis of the dipole does not coincide with the axis of rotation of the earth but is presently tilted by approxmately $11.3^{\circ}$ with respect to the later. This results into two situations as given in the figure ahead.

Hence, the declination varies between $11.3^{\circ} \mathrm{W}$ to $11.3^{\circ} \mathrm{E}$.

Thinking Process

Permanent magnet at room temperature behave as a ferromagnetic substance for a long period of time.

Answer

(d) As we know a permanent magnet is a substance which at room temperature retain ferromagnetic property for a long period of time.

The individual atoms in a ferromagnetic material possess a dipole moment as in a paramagnetic material.

However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. Thus, we can say that in a permanent magnet at room temperature, domains are all perfectly aligned.

Thinking Process

The electric field lines, do not form a continuous closed path while the magnetic field lines form the closed paths.

Answer

(b) As Gauss’ law states, $\int_{S} E \cdot d s=\frac{q}{\varepsilon_{0}}$ for electrostatic field. It does not contradict for electrostatic fields as the electric field lines do not form continuous closed path.

According to Gauss’ law in magnetic field,

$$ \int_{S} \mathrm{E} \cdot \mathrm{ds}=0 $$

It contradicts for magnetic field, because there is a magnetic field inside the solenoid and no field outside the solenoid carrying current but the magnetic field lines form the closed path.

Thinking Process

From Curie law, we know that magnetisation is directly proportional to the magnetic field induction and inversely proportional to the temperature in kelvin.

Answer

(b) As Curie law explains, we can deduce a formula for the relation between magnetic field induction, temperature and magnetisation.

i.e., $\quad I$ (magnetisation $) \propto \frac{B \text { (magnetic field induction })}{t \text { (temperature in kelvin })}$

$\begin{array}{rlrl} \Rightarrow & \frac{I_2}{I_1} =\frac{B_2}{B_1} \times \frac{t_1}{t_2} \\ \text { Let us suppose, here } & I_1 =8 \mathrm{Am}^{-1} \\ B_1 & =0.6 \mathrm{~T}, t_1=4 \mathrm{~K} \\ B_2 & =0.2 \mathrm{~T}, t_2=16 \mathrm{~K} \\ I_2=? \\ \Rightarrow \quad & \frac{0.2}{0.6} \times \frac{4}{16} =\frac{I_2}{8} \\ \Rightarrow \quad & I_2 =8 \times \frac{1}{12}=\frac{2}{3} \mathrm{Am}^{-1} \end{array}$

Thinking Process

According to the properties of magnetic field lines (B), for any magnet, it forms continuous closed loops. This is unlike the electric dipole where these field lines begin from a positive charge and end on the negative charge or escape to infinity.

Also, magnetic intensity $(H)$ outside any magnet is $H=B / \mu_{0}$ and for inside the magnet $H=B / \mu_{0} \mu_{r}$, where $\mu_{r}$ is the relative permeability of material (magnetic).

Answer

$(a, d)$

Magnetic field lines for magnetic induction (B) form continuous lines. So, lines of B are necessarily continuous across $S$.

Also, magnetic intensity $\mathbf{( H )}$ varies for inside and outside the lump. So, lines of $\mathbf{H}$ cannot all be continuous across $S$.

Answer

$(a, d)$

The primary origin of magnetism lies in the fact that the electrons are revolving and spinning about nucleus of an atom, which gives rise to current called atomic current.

This atomic currents gives rise to magnetism. The revolving and spinning about nucleus of an atom is called intrinsic spin of electron.

Thinking Process

The magnetic intensity $\mathbf{H}$ field $=n \boldsymbol{I}$, where $n=$ number of turns per metre of a solenoid

and $I=$ current and $B=\mu_{0} \mu_{r} n I$.

Also, at normal temperature, a solenoid behave as a ferromagnetic substand and at the temperature beyond the Curie temperature, it behaves as a paramagnetic substance.

Answer

(a, $d)$

Here, for solenoid $\mathbf{H}=n I$.

$\Rightarrow \quad \mathrm{H}=1000 \times 1=1000 \mathrm{Am}$

Thus, $\mathbf{H}$ is a constant, so it is nearly unchanged.

But

$$ \begin{aligned} B & =\mu_{0} \mu_{r} n I \\ & =\mu_{0} n I \mu_{r} \\ & =k \text { (constant) } \mu_{r} . \end{aligned} $$

Thus, from above equation, we find that $\mathbf{B}$ varies with the variation in $\mu_{r}$.

Now, for magnetisation in the core, when temperature of the iron core of solenoid is raised beyond Curie temperature, then it behave as paramagnetic material, where

$\begin{aligned} \text{and} \quad (\chi_m) _{\text {Fero }} & \approx 10^3 \\ \text{and} \quad (\chi_m) _{\text {Para }} & \approx 10^{-5} \\ \Rightarrow \quad\frac{(\chi_mt) _{\text {Fero }}}{(\chi_m)} & =\frac{10^3}{10^{-5}}=10^8 \end{aligned}$

Answer

$(a, c, d)$

Electrostatic shielding is the phenomenon to block the effects of an electric field. The conducting shell can block the effects of an external field on its internal content or the effect of an internal field on the outside environment.

Magnetostatic shielding is done by using an enclosure made of a high permeability magnetic material to prevent a static magnetic field outside the enclosure from reaching objects inside it or to confine a magnetic field within the enclosure.

Thinking Process

Angle of inclination or dip is the angle that the total magnetic field of the earth makes with the surface of the earth.

Answer

$(b, c, d)$

If the total magnetic field of the earth is modelled by a point magnetic dipole at the centre, then it is in the same plane of geographical equator, thus the angle of dip at a point on the geographical equator is bounded in a range from positive to negative value.

Thinking Process

Mass of a proton is very larger than the mass of an electron, so its spinning is negligible as compared to that of electron spin.

Answer

The comparison between the spinning of a proton and an electron can be done by comparing their magnetic dipole moment which can be given by

$$ \begin{aligned} & M & =\frac{e h}{4 \pi m} \text { or } M \propto \frac{1}{m} & \newline \ \therefore & & \frac{M_{p}}{M_{e}} & =\frac{m_{e}}{m_{p}} \ & =\frac{M_{e}}{1837 M_{e}} & & \left(\because \frac{e h}{4 \pi}=\text { constant }\right) \ \Rightarrow & & \frac{M_{p}}{M_{e}} & =\frac{1}{1837} \ll 1 \ \Rightarrow & M_{p} & \ll M_{e} & \left(\because M_{p}=1837 m_{e}\right) \end{aligned} $$

Thus, effect of magnetic moment of proton is neglected as compared to that of electron.

Answer

Given, $M$ (intensity of magnetisation) $=10^{6} \mathrm{~A} / \mathrm{m}$.

$$ \begin{aligned} l \text { (length) } & =10 \mathrm{~cm}=10 \times 10^{-2} \mathrm{~m}=0.1 \mathrm{~m} \\ I_{M} & =\text { magnetisation current } \\ M & =\frac{I_{M}}{l} \\ I_{M} & =M \times l \\ & =10^{6} \times 0.1=10^{5} \mathrm{~A} \end{aligned} $$

$$ \text { and } \quad I_{M}=\text { magnetisation current } $$

We know that

Note Here, $M=$ intensity of magnetisation as its unit is given as $A / m$.

Thinking Process

Magnetic susceptibility is a measure of how a magnetic material responds to an external field.

Answer

We know that

Also,

$$ \begin{array}{r} \text { Density of nitrogen } \rho_{N}{2}=\frac{28 ~g}{22.4 ~L}=\frac{28 ~g}{22400 cc} \\ \text { density of copper } \rho_{Cu}=\frac{8 ~g}{22.4 L}=\frac{8 ~g}{22400 cc} \end{array} $$

Now, comparing both densities

Also given

$$ \begin{aligned} \frac{\rho_{N_2}}{\rho_{{Cu}}} & =\frac{28}{22400} \times \frac{1}{8}=1.6 \times 10^{-4} \newline \ \frac{\chi_{N_{2}}}{\chi_{{Cu}}} & =\frac{5 \times 10^{-9}}{10^{-5}}=5 \times 10^{-4} \newline \ \chi & =\frac{\text { Magnetisation }(M)}{\text { Magnetic intensity }(H)} \newline \ & =\frac{\text { Magnetic moment }(M) / \text { Volume }(V)}{H} \newline \ & =\frac{M}{H V}=\frac{M}{H \text { (mass / density) }}=\frac{M \rho}{H m} \end{aligned} $$

We know that, $\quad \chi=\frac{\text { Magnetisation }(M)}{\text { Magnetic intensity }(H)}$

$\therefore \quad \chi \propto \rho \quad\left(\because \frac{M}{H m}=\right.$ constant $)$

Hence,

$$ \frac{{N_2}}{{{Cu}}}=\frac{\rho_{N_2}}{{Cu}}=1.6 \times 10^{-4} $$

Thus, we can say that magnitude difference or major difference between the diamagnetic susceptibility of ${N}_{2}$ and ${Cu}$.

Answer

Susceptibility of magnetic material $\chi=\frac{I}{H}$, where $I$ is the intensity of magnetisation induced in the material and $H$ is the magnetising force.

Diamagnetism is due to orbital motion of electrons in an atom developing magnetic moments opposite to applied field. Thus, the resultant magnetic moment of the diamagnetic material is zero and hence, the susceptibility $\chi$ of diamagnetic material is not much affected by temperature.

Paramagnetism and ferromagnetism is due to alignments of atomic magnetic moments in the direction of the applied field. As temperature is raised, the alignment is disturbed, resulting decrease in susceptibility of both with increase in temperature.

Thinking Process

A superconducting material and nitrogen are diamagnetic in nature.

Answer

When a diamagnetic material is dipped in liquid nitrogen, it again behaves as a diamagnetic material. Thus, superconducting material will again behave as a diamagnetic material. When this diamagnetic material is placed near a bar magnet, it will be feebly magnetised opposite to the direction of magnetising field.

(i) Thus, it will be repelled.

(ii) Also its direction of magnetic moment will be opposite to the direction of magnetic field of magnet.

Answer

Let us draw the figure for given situation,

We have to prove that $g \mathbf{B} \cdot d \mathbf{S}=0$. This is called Gauss’s law in magnetisation.

According to question,

Magnetic moment of dipole at origin $O$ is

$$ \mathbf{M}=M \hat{k} $$

Let $P$ be a point at distance $r$ from $O$ and $O P$ makes an angle $\theta$ with $z$-axis. Component of $\mathbf{M}$ along $O P=M \cos \theta$.

Now, the magnetic field induction at $P$ due to dipole of moment $\mathbf{M} \cos \theta$ is

$$ \mathbf{B}=\frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{r} $$

From the diagram, $r$ is the radius of sphere with centre at $O$ lying in yz-plane. Take an elementary area $\mathbf{S}$ of the surface at $P$. Then,

$$ \begin{aligned} d \mathbf{S} & =r(r \sin \theta d \theta) \hat{\mathbf{r}}=r^{2} \sin \theta d \theta \hat{\mathbf{r}} \\ g \mathbf{B} \cdot d \mathbf{S} & =\oint \frac{\mu_{0}}{4 \pi} \frac{2 M \cos \theta}{r^{3}} \hat{\mathbf{r}}\left(r^{2} \sin \theta d \theta \hat{\mathbf{r}}\right) \\ & =\frac{\mu_{0}}{4 \pi} \frac{M}{r} \int_{0}^{2 \pi} 2 \sin \theta \cdot \cos \theta d \theta \\ & =\frac{\mu_{0}}{4 \pi} \frac{M}{r} \int_{0}^{2 \pi} \sin 2 \theta d \theta \\ & =\frac{\mu_{0}}{4 \pi} \frac{M}{r} \frac{-\cos 2 \theta}{2} 0 \\ & =-\frac{\mu_{0}}{4 \pi} \frac{M}{2 r}[\cos 4 \pi-\cos 0] \\ & =\frac{\mu_{0}}{4 \pi} \frac{M}{2 r}[1-1]=0 \end{aligned} $$

Answer

The system will be in stable equilibrium if the net force on the system is zero and net torque on the system is also zero. This is possible only when the poles of the remaining two magnets are as given in the figure.

Thinking Process

$E(r)=c B(r)$, suppose the angle between $\mathbf{p}$ and $\mathbf{E}$ is $\boldsymbol{\theta}$. Torque on electric dipole of moment

$\mathbf{p}$ in electric field $\mathbf{E}, \tau=p E \sin \theta$.

Answer

Now, suppose that the angle between $\mathbf{M}$ and $B$ is $\theta$.

Torque on magnetic dipole moment $\mathbf{M}$ in magnetic field $\mathbf{B}$,

$$ \tau^{\prime}=M B \sin \theta $$

Two motions will be identical, if

$$ \Rightarrow $$

But,

$$ \begin{aligned} p E \sin \theta & =M B \sin \theta \ p E & =M B \ E & =c B \end{aligned} $$

$\therefore$ Putting this value in E(i),

$$ \begin{gathered} p c B=M B \ p=\frac{M}{c} \end{gathered} $$

Answer

Given, $I=$ moment of inertia of the bar magnet

$m=$ mass of bar magnet

$l=$ length of magnet about an any passing through its centre and perpendicular to its length

$M=$ magnetic moment of the magnet

$B=$ uniform magnetic field in which magnet is oscillating, we get time period of

oscillation is,

$$ T=2 \pi \sqrt{\frac{I}{M B}} $$

Here,

$$ I=\frac{m l^{2}}{12} $$

When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is

Magnetic dipole moment

$$ I^{\prime}=\frac{m}{2} \frac{(l / 2)^{2}}{12}=\frac{m l^{2}}{12} \times \frac{1}{8}=\frac{I}{8} $$

Its time period of oscillation is

$$ M^{\prime}=M / 2 $$

$$ \begin{aligned} & T^{\prime}=2 \pi \sqrt{\frac{I^{\prime}}{M^{\prime} B}}=2 \pi \sqrt{\frac{I / 8}{(M / 2) B}}=\frac{2 \pi}{2} \sqrt{\frac{I}{M B}} \\ & T^{\prime}=\frac{T}{2} \end{aligned} $$

Answer

Consider a magnetic field line of $\mathbf{B}$ through the bar magnet as given in the figure below.

The magnetic field line of $B$ through the bar magnet must be a closed loop.

Let $C$ be the amperian loop. Then,

$$ \int_{Q}^{P} H \cdot d l=\int_{Q}^{P} \frac{B}{m_{0}} \cdot d l $$

We know that the angle between $\mathbf{B}$ and $\mathbf{d I}$ is less than $90^{\circ}$ inside the bar magnet. So, it is positive.

i.e.,

$$ \int_{Q}^{P} \mathrm{H} \cdot \mathrm{dl}=\int_{Q}^{P} \frac{\mathrm{B}}{\mu_{0}} \cdot \mathrm{dl}>0 $$

Hence, the lines of $\mathrm{B}$ must run from south pole $(S)$ to north pole $(N)$ inside the bar magnet.

According to Ampere’s law,

$$ \begin{array}{ll} \therefore & \oint_{P Q P} \mathrm{H} \cdot \mathrm{dl}=0 \\ \therefore & \oint_{P Q P} \mathrm{H} \cdot \mathrm{dl}=\int_{P}^{Q} \mathrm{H} \cdot \mathrm{dl}+\int_{Q}^{P} \mathrm{H} \cdot \mathrm{dl}=0 \\ \text { As } & \int_{Q}^{P} \mathrm{H} . \mathrm{dl}>0, \mathrm{so}, \int_{P}^{Q} \mathrm{H} \cdot \mathrm{dl}<0 \end{array} $$

It will be so if angle between $\mathbf{H}$ and $\mathbf{~ d l}$ is more than $90^{\circ}$, so that $\cos \theta$ is negative. It means the line of $\mathrm{H}$ must run from $\mathrm{N}$-pole to $\mathrm{S}$-pole inside the bar magnet.

Thinking Process

Let us consider the figure below

Answer

From $P$ to $Q$, every point on the $z$-axis lies at the axial line of magnetic dipole of moment $\mathbf{M}$. Magnetic field induction at a point distance $z$ from the magnetic dipole of moment is

$$ |\mathbf{B}|=\frac{\mu_{0}}{4 \pi} \frac{2|\mathbf{M}|}{z^{3}}=\frac{\mu_{0} M}{2 \pi z^{3}} $$

(i) Along z-axis from $P$ to $Q$.

$$ \begin{aligned} \int_{\rho}^{Q} B \cdot d l & =\int_{\rho}^{Q} B \cdot d l \cos 0^{\circ}=\int_{a}^{B} B d z \\ & =\int_{a}^{R} \frac{\mu_{0}}{2 \pi} \frac{M}{z^{3}} d z=\frac{\mu_{0} M}{2 \pi} \frac{-1}{2} \frac{1}{R^{2}}-\frac{1}{a^{2}} \\ & =\frac{\mu_{0} M}{4 \pi} \frac{1}{a^{2}}-\frac{1}{R^{2}} \end{aligned} $$

(ii) Along the quarter circle $Q S$ of radius $R$ as given in the figure below

The point $A$ lies on the equatorial line of the magnetic dipole of moment $M \sin \theta$. Magnetic field at point $A$ on the circular arc is

$$ \begin{aligned} B & =\frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{R^{3}} ; \mathrm{dl}=R d \theta \\ \therefore \quad & \int \mathrm{B} \cdot \mathrm{dl}=\int \mathrm{B} \cdot \mathrm{dl} \cos \theta=\int_{0}^{\frac{\pi}{2}} \frac{\mu_{0}}{4 \pi} \frac{M \sin \theta}{R^{3}} R d \theta \end{aligned} $$

Circular arc $=\frac{\mu_0}{4 \pi} \frac{M}{R}(-\cos \theta)_{0}^{\pi / 2}=\frac{\mu_0}{4 \pi} \frac{M}{R^{2}}$

(iii) Along $x$-axis over the path $S T$, consider the figure given ahead

From figure, every point lies on the equatorial line of magnetic dipole. Magnetic field induction at a point distance $x$ from the dipole is

$$ \begin{aligned} & B=\frac{\mu_{0}}{4 \pi} \frac{M}{x^{3}} \\ & \therefore \quad \int_{\beta}^{T} \cdot \mathbf{d l}=\int_{R}^{a}-\frac{\mu_{0} \mathbf{M}}{4 \pi x^{3}} \cdot d \mathbf{d l}=0\left[\because \text { angle between }(-\mathbf{M}) \text { and } \mathbf{d l} \text { is } 90^{\circ}\right] \end{aligned} $$

(iv) Along the quarter circle TP of radius a. Consider the figure given below

From case (ii), we get line integral of $\mathbf{B}$ along the quarter circle $T P$ of radius $a$ is circular arc TP

$$ \begin{aligned} & \int B \cdot dl \int_{\pi / 2}^0 \frac{\mu_0}{4 \pi} \frac{M \sin \theta}{a^{3}} ad \theta \\ & =\frac{\mu_0}{4 \pi} \frac{M}{a^{2}} \int_{\pi / 2}^0 \sin \theta d \theta=\frac{\mu_0}{4 \pi} \frac{M}{a^{2}}[-\cos \theta]_{\pi / 2}^0 \\ & =\frac{-\mu_0}{4 \pi} \frac{M}{a^{2}} \\ & =\frac{\mu_0 M}{4} \frac{1}{a^{2}}-\frac{1}{R^{2}}+\frac{\mu_0}{4 \pi} \frac{M}{R^{2}}+0+-\frac{\mu_0}{4 \pi} \frac{M}{a^{2}}=0 \end{aligned} $$

$$ \therefore \quad \oint_{P Q S T} B \cdot d l=\int_{P}^{Q} B \cdot d l+\int_{Q}^{S} B \cdot d l+\int_{S}^{T} B \cdot d l+\int_{T}^{P} B \cdot d l $$

Thinking Process

Magnetic susceptibility is a measure of how a magnetic material responds to an external field. i.e., magnetic susceptibility

$$ \chi_{m}=\frac{I}{H}=\frac{(\text { Intensity of magnetisation) }}{\text { (Magnetising force) }} $$

Answer

As $I$ and $H$ both have same units and dimensions, hence, $\chi$ has no dimensions. Here, in this question, $\chi$ is to be related with $e, m, v, R$ and $\mu_{0}$. We know that dimensions of $\mu_{0}=\left[\mathrm{ML} \theta^{-2}\right]$

From Biot-Savart’s law,

$$ \begin{aligned} d B & =\frac{\mu_{0}}{4 \pi} \frac{I d l \sin \theta}{r^{2}} \\ \Rightarrow \quad \mu_{0} & =\frac{4 \pi r^{2} d B}{I d l \sin \theta}=\frac{4 \pi r^{2}}{I d l \sin \theta} \times \frac{f}{q v \sin \theta} \quad \because d B=\frac{F}{q v \sin \theta} \\ \therefore \quad \text { Dimensions of } \mu_{0} & =\frac{L^{2} \times\left(\mathrm{MLT}^{-2}\right)}{\left(\mathrm{QT}^{-1}\right)(\mathrm{L}) \times 1 \times(\mathrm{Q})\left(\mathrm{LT}^{-1}\right) \times(1)} \newline=\left[\mathrm{MLQ}^{-2}\right] \end{aligned} $$

where $Q$ is the dimension of charge.

As $\chi$ is dimensionless, it should have no involvement of charge $Q$ in its dimensional formula. It will be so if $\mu_{0}$ and $e$ together should have the value $\mu_{0} e^{2}$, as $e$ has the dimensions of charge.

Let

$$ \chi=\mu_{0} e^{2} m^{a} v^{b} R^{c} $$

where $a, b, c$ are the power of $m, v$ and $R$ respectively, such that relation (i) is satisfied.

Dimensional equation of (i) is

$$ \begin{aligned} {\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \mathrm{Q}^{0}\right] } & =\left[\mathrm{ML} \mathrm{Q}^{-2}\right] \times\left[\mathrm{Q}^{2}\right]\left[\mathrm{M}^{a}\right] \times\left(\mathrm{LT}^{-1}\right)^{b} \times[\mathrm{L}]^{c} \\ & =\left[\mathrm{M}^{1+a}+\mathrm{L}^{1+b+c} \mathrm{~T}^{-b} \mathrm{Q}^{0}\right] \end{aligned} $$

Equating the powers of $M, L$ and $T$, we get

$$ \begin{aligned} & 0=1+a \Rightarrow a=-1,0=1+b+c \\ & 0=-b \Rightarrow b=0,0=1+0+c \text { or } c=-1 \end{aligned} $$

Putting values in Eq. (i), we get

$$ \begin{aligned} \chi & =\mu_{0} e^{2} m^{-1} v^{2} R^{-1}=\frac{\mu_{0} e^{2}}{m R} \\ \mu_{0} & =4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}{ }^{-1} \\ e & =1.6 \times 10^{-19} \mathrm{C} \\ m & =9.1 \times 10^{-31} \mathrm{~kg}, R=10^{-10} \mathrm{~m} \\ \chi & =\frac{\left(4 \pi \times 10^{-7}\right) \times\left(1.6 \times 10^{-19}\right)^{2}}{\left(9.1 \times 10^{-31}\right) \times 10^{-10}} \approx 10^{-4} \end{aligned} $$

$$ \therefore \quad \frac{\chi}{\chi_{\text {(given solid) }}}=\frac{10^{-4}}{10^{-5}}=10 $$

Answer

(a)

$$ \begin{aligned} B_{V} & =\frac{\mu_{0}}{4 \pi} \frac{2 m \cos \theta}{r^{3}} \\ B_{H} & =\frac{\mu_{0}}{4 \pi} \frac{\sin \theta m}{r^{3}} \end{aligned} $$

Squaring both the equations and adding, we get

$$ \begin{aligned} B_{V}^{2}+B_{H}^{2} & =\frac{\mu_{0}}{4 \pi} \frac{m^{2}}{r^{6}}\left[4 \cos ^{2} \theta+\sin ^{2} \theta\right] \\ B & =\sqrt{B_{V}^{2}+B_{H}^{2}}=\frac{\mu_{0}}{4 \pi} \frac{m}{r^{3}}\left[3 \cos ^{2} \theta+1\right]^{1 / 2} \end{aligned} $$

From Eq. (iii), the value of $B$ is minimum, if $\cos \theta=\frac{\pi}{2}$

$\theta=\frac{\pi}{2}$. Thus, the magnetic equator is the locus. (b) Angle of dip,

$$ \begin{aligned} & \tan \delta=\frac{B_{V}}{B_{H}}=\frac{\frac{\mu_{0}}{4 \pi} \cdot \frac{2 m \cos \theta}{r^{3}}}{\frac{\mu_{0}}{4 \pi} \cdot \frac{\sin \theta \cdot m}{r^{3}}}=2 \cot \theta \\ & \tan \delta=2 \cot \theta \end{aligned} $$

For dip angle is zero i.e., $\delta=0$

$$ \begin{aligned} \cot \theta & =0 \\ \theta & =\frac{\pi}{2} \end{aligned} $$

It means that locus is again magnetic equator.

(c) $\tan \delta=\frac{B_{V}}{B_{H}}$

Angle of dip i.e., $\delta= \pm 45$

$$ \begin{aligned} \frac{B_{V}}{B_{H}} & =\tan \left( \pm 45^{\circ}\right) \\ \frac{B_{V}}{B_{H}} & =1 \\ 2 \cot \theta & =1 \\ \cot \theta & =\frac{1}{2} \\ \Rightarrow \quad \tan \theta & =2 \\ \theta \quad \theta & =\tan ^{-1}(2) \end{aligned} $$

Thus, $\theta=\tan ^{-1}(2)$ is the locus.

Answer

$P$ is in the plane $S$, needle is in north, so the declination is zero.

$P$ is also on the magnetic equator, so the angle of dip $=0$, because the value of angle of dip at equator is zero. $Q$ is also on the magnetic equator, thus the angle of dip is zero. As earth tilted on its axis by $11.3^{\circ}$, thus the declination at $Q$ is $11.3^{\circ}$.

Answer

$C_{1}=$ circular coil of radius $R$, length $L$, number of turns per unit length

$$ n_{1}=\frac{L}{2 \pi R} $$

$C_{2}=$ square of side $a$ and perimeter $L$, number of turns per unit length $n_{2}=\frac{L}{4 a}$

Magnetic moment of $C_{1}$

$\Rightarrow$ Magnetic moment of $C_{2}$

$$ m_{1}=n_{1} I A_{1} $$

$\Rightarrow$

$$ \begin{aligned} m_{2} & =n_{2} I A_{2} \\ m_{1} & =\frac{L \cdot I \cdot \pi R^{2}}{2 \pi R} \\ m_{2} & =\frac{L}{4 a} \cdot I \cdot a^{2} \\ m_{1} & =\frac{L I R}{2} \\ m_{2} & =\frac{L I a}{4} \end{aligned} $$

Moment of inertia of $C_{1} \Rightarrow I_{1}=\frac{M R^{2}}{2}$

Moment of inertia of $C_{2} \Rightarrow I_{2}=\frac{M a^{2}}{12}$

Frequency of $C_{1} \Rightarrow t_{1}=2 \pi \sqrt{\frac{I_{1}}{m_{1} B}}$

Frequency of $C_{2} \Rightarrow f_{2}=2 \pi \sqrt{\frac{I_{2}}{m_{2} B}}$

According to question, $f_{1}=f_{2}$

$$ \begin{aligned} 2 \pi \sqrt{\frac{I_{1}}{m_{1} B}} & =2 \pi \sqrt{\frac{I_{2}}{m_{2} B}} \\ \frac{I_{1}}{m_{1}} & =\frac{I_{2}}{m_{2}} \text { or } \frac{m_{2}}{m_{1}}=\frac{I_{2}}{I_{1}} \end{aligned} $$

Plugging the values by Eqs. (i), (ii), (iii) and (iv)

$$ \begin{aligned} \frac{L I a \cdot 2}{4 \times L I R} & =\frac{M a^{2} \cdot 2}{12 \cdot M R^{2}} \\ \frac{a}{2 R} & =\frac{a^{2}}{6 R^{2}} \\ 3 R & =a \end{aligned} $$

Thus, the value of $a$ is $3 R$.