Question

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 :

• A. $2\mu V$
• B. $1\mu V$
• C. $0.75\mu V$
• D. $0.5\mu V$

Answer 1

Answer: (B)

In given $fi{g}_x=15$ because left arm of the frame is at $10cm$ from the wire. and $a=10cm$. emf in
$AD⟹e_1⟹\frac{a{\mu }_{o}iv}{2\pi (x-a/2)}$
emf in $EF⟹e_2⟹\frac{a{\mu }_{o}iv}{2\pi (x+a/2)}$
Net emf $=e_1-e_2=\frac{a{\mu }_{o}iv}{2\pi (x-a/2)}-\frac{a{\mu }_{o}iv}{2\pi (x+a/2)}$
putting the given values into the equation we get, emf induced $=1.3\mu V\approx 1\mu V$