# (a) Define the term ‘self inductance’ of a coil. Write its S.I. unit.(b)  A rectangular loop of sides a and b carrying current $I_{2}$ is kept at a distance ‘a’ from an infinitely long straight wire carrying  current $I_{1}$ as shown in the figure. Obtain an expression for the resultant force acting on the loop.

Self-inductance (L) is the emf induced in the coil when the rate of change of current in the coil is $1$ ampere/second

$\phi =LI$

Where $\phi$= magnetic flux

$I$= Current flowing

Where $L$= coefficient of self- inductance or self-inductance of the coil

when, $L=1A$

Then, $\phi =L$

The S.I unit of self-inductance is Henry (H).

(b) We know the magnetic field due to the infinitely long current-carrying wire is given by :

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

Where, $\mu _{0}=4\pi \times 10^{-7}H/m$

$i$ = current carried by infinite wire

$r$= perpendicular distance of wire to that point.

Now, the force acting on a straight wire is given by,

$F=i(l\times B)$

For side  AB

$\vec{F}=i(\vec{l}\times \vec{B})$

$F=I_{2}\left ( b\times \frac{\mu _{0}I_{1}}{2\pi a} \right )$

$F=\frac{I_{1}I_{2}\mu _{0}b}{2\pi a},$

in -ve x-direction.

Such that, for BC and AD forces are equal but opposite in direction so, they will cancel out each other.

Now, for side CD

$\vec{F}=i(\vec{l}\times \vec{B})$

$F=I_{2}\left ( b\frac{\mu _{0}I_{1}}{2\pi (2a)} \right )$

$F=\frac{I_{1}I_{2}\mu _{0}b}{4\pi a}$ in +ve x- direction

Hence, the net force will be in -ve x-direction ,

$F_{net}=\frac{I_{1}I_{2}\mu _{0}b}{\pi a}\left ( \frac{1}{2}-\frac{1}{4} \right )$

$F_{net}=\frac{I_{1}I_{2}\mu _{0}b}{4\pi a}$

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