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 $\mathrm{O}$

$$\begin{array}{rlr}\overrightarrow{B_2}& =\frac{{\mu }_0}{4\pi }\cdot \frac{3x}{2}\cdot \frac{I_2}{R}& =\frac{3{\mu }_0}{8}\frac{I_2}{R}\end{array}$$

(in the direction of $B_2$ )

Netmagnetic field, $\overrightarrow{B}=\overrightarrow{B_1}+\overrightarrow{B_2}$

$$\begin{array}{}\text{(i)}& |\overrightarrow{B}|=\frac{{\mu }_0{I}_1}{8R}-\frac{{\mu }_0{I}_2}{8R}\times 3\end{array}$$

But $I_1=3I_2$ (As resistance of bigger part is three times that of the smaller part of the loop.)

Substituting $I_1=3I_2$ in equation (i),

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

$1/2$

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,

$$\oint B\cdot dI={\mu }_{0}I$$

where,

$B=$ magnetic field

$dl=$ 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,

$$B={\mu }_{0}nI$$

• at the ends of solenoid, $B_{\text{end }}=\frac{{\mu }_{0}nI}{2}$

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 :

$$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}i\Rightarrow B\times 2\pi \mathrm{r}={\mu }_{0}Ni$$

where, $r=$ average radius

Here,

$$B=\frac{{\mu }_{0}Ni}{2\pi r}={\mu }_{0}ni$$

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

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

$$\oint \overrightarrow{B}\cdot \overrightarrow{dl}={\mu }_{0}I$$

• Magnetic field at the surface of a solid cylinder :

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

(Magnetic field inside the solenoid :

$$B={\mu }_{0}nI$$

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

$$r=\frac{{\mu }_{0}Ni}{2\pi r}$$

Force between two parallel current carrying wires : $F=\frac{{\mu }_0}{2\pi }{I}_1{I}_{2}L$

? Objective Type Questions