Solution:

Ans. (i) (a) Try yourself, Similar to Q. 3 (b), Short Answer Type Questions-II.

(b) Try yourself, Similar to Q. 1 (ii), Short Answer Type Questions-I

(ii) Let current $I$ be divided at point $M$ into two parts $I_{1}$ and $I_{2}$; in bigger and smaller parts of the loop respectively.

Magnetic field of current $I_{1}$ at point $O$,

$$ \overrightarrow{B_{1}}=\frac{\mu_{0}}{4 \pi} \frac{\theta I_{1}}{R}=\frac{\mu_{0}}{4 \pi} \times \frac{\pi}{2} \times \frac{I_{1}}{r} $$

$$ =\frac{\mu_{0}}{8} \frac{I_{1}}{R} $$

Magnetic field due to current $\mathrm{I}_{2}$ at point $\mathrm{O}$

$$ \begin{aligned} \overrightarrow{B_{2}} & =\frac{\mu_{0}}{4 \pi} \cdot \frac{3 x}{2} \cdot \frac{I_{2}}{R} \ & =\frac{3 \mu_{0}}{8} \frac{I_{2}}{R} \end{aligned} $$

(in the direction of $B_{2}$ )

Netmagnetic field, $\vec{B}=\overrightarrow{B_{1}}+\overrightarrow{B_{2}}$

$$ \begin{equation*} |\vec{B}|=\frac{\mu_{0} I_{1}}{8 R}-\frac{\mu_{0} I_{2}}{8 R} \times 3 \tag{i} \end{equation*} $$

But $I_{1}=3 I_{2}$ (As resistance of bigger part is three times that of the smaller part of the loop.)

Substituting $I_{1}=3 I_{2}$ in equation (i),

$$ |\vec{B}|=0 $$

$1 / 2$

10

TOPIC-2

Ampere’s Circuital Law and its Applications

Revision Notes

• Ampere’s circuital law states that the line integral of magnetic field around a closed path is $\mu_{0}$ times of total current enclosed by the path,

$$ \oint B \cdot d I=\mu_{0} I $$

where,

$B=$ magnetic field

$d l=$ infinitesimal segment of the path

$\mu_{0}=$ permeability of free space

$I=$ enclosed electric current by the path

• Magnetic field at a point will not depend on the shape of Amperian loop and will remain same at every point on the loop.

Forces between two parallel currents

• Two parallel wires separated by distance $r$ having currents $I_{1}$ and $I_{2}$ where magnetic field strength at second wire due to current flowing in first wire is given as :

$$ B=\frac{\mu_{0} I_{1}}{2 \pi r} $$

• In this, the field is orientated at right-angles to second wire where force per unit length on the second wire will be :

$$ \frac{F}{l}=\frac{\mu_{0} I_{1} I_{2}}{2 \pi r} $$

• Magnetic field-strength at first wire due to the current flowing in second wire will be :

$$ B=\frac{\mu_{0} I_{2}}{2 \pi r} $$

One ampere is the magnitude of current which, when flowing in each parallel wire one metre apart, results in a force between the wires as $2 \times 10^{-7} \mathrm{~N}$ per meter of length.

Applications of Ampere’s law to infinitely long straight wire, straight and toroidal solenoids :

(i) Magnetic Field due to long straight wire

Amperes law describes the magnitude of magnetic field of a straight wire as :

$$ B=\frac{\mu_{0} I}{2 \pi r} $$

where,

• Field $B$ is tangential to a circle of radius $r$ centeredôn the wire.
• Magnetic field $B$ and path length $L$ will remain parallet where magnetic field travels.

(ii) Magnetic Field due to Solenoid

Solenoid : An electromagnet that generates a controlled magnetic field.

Solenoid is a tightly wound helical coil ofyire whose diameter is small compared to its length.

• Magnetic field generated in the centre, or core of a current carrying solenoid is uniform and is directed along the axis of solenoid.
• Magnetic field due to a straight solenoid:
• at any point in the solenoid,

$$ B=\mu_{0} n I $$

• at the ends of solenoid, $\quad B_{\text {end }}=\frac{\mu_{0} n I}{2}$

where, $n=$ number of turns per unit length, $I=$ current in the coil.

(iii) Magnetic Field due to Toroid

Toroid : It is an electronic component made of hollow circular ring wound with number of turns of copper wire.

The toroid is a hollow circular ring on which a large number of turns of a wire are closely wound.

• A toroid with $n$ turns per unit length with mean radius $r$, where current $i$ is flowing through it, then the magnetic field experienced by the toroid with total number of turns $\mathrm{N}$ will be :

$$ \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} i \Rightarrow B \times 2 \pi \mathrm{r}=\mu_{0} N i $$

where, $r=$ average radius

Here,

$$ B=\frac{\mu_{0} N i}{2 \pi r}=\mu_{0} n i $$

$$ \left(\text { here }, n=\frac{N}{2 \pi r}\right) $$

At any point, empty space surrounded by toroid and outside the toroid, magnetic field B will be zero as net current is zero.

Key Formulae

Ampere’s circuital law

$$ \oint \vec{B} \cdot \overrightarrow{d l}=\mu_{0} I $$

• Magnetic field at the surface of a solid cylinder :

$$ \mathrm{B}=\frac{\mu_{0} I}{2 \pi R} $$

(Magnetic field inside the solenoid :

$$ B=\mu_{0} n I $$

D Magnetic field in a toroid with mean radius $r$ :

$$ r=\frac{\mu_{0} N i}{2 \pi r} $$

Force between two parallel current carrying wires : $F=\frac{\mu_{0}}{2 \pi} I_{1} I_{2} L$

? Objective Type Questions