# Find induced emf

1 vote
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A square frame of side 10 cm and a long straight
wire carrying current 1 A are in the plane of the
paper. Starting from close to the wire, the frame
moves towards the right with a constant speed of
10 ms–1 (see figure). The e.m.f induced at the time
the left arm of the frame is at x = 10 cm from the
wire is :

I tried as

But answer is given as $2\mu V$

1 vote

There is no need to integrate and then differentiate.

The magnetic fields at the LH and RH vertical sides of the loop are $B_1=\frac{\mu_o I}{2\pi x}$ and $B_2=\frac{\mu_o I}{2\pi (x+l)}$.

At the LHS the flux through the loop is decreasing at the rate $B_1 lv$ and at the RHS it is increasing at the rate $B_2 lv$. The rate of change of flux is the induced emf :
$V=-\frac{d\phi}{dt}=-(B_2-B_1)lv=\frac{\mu_o I lv}{2\pi}(\frac{1}{x}-\frac{1}{x+l})=\frac{\mu_o I lv}{2\pi}\frac{l}{x(x+l)}$.

Putting in values we get
$V=\frac{4\pi\times 10^{-7}\times 1\times 0.1\times 10\times 0.1}{2\pi \times0.1\times 0.2}=10^{-6}V$.