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 $fig _{ x }=15$ because left arm of the frame is at $10 cm$ from the wire. and $a = 1 0 cm$. emf in
$A D \Longrightarrow e _{1} \Longrightarrow \frac{ a \mu_{ o } iv }{2 \pi( x - a / 2)}$
emf in $E F \Longrightarrow e _{2} \Longrightarrow \frac{ a \mu_{ o } i v }{2 \pi( x + a / 2 )}$
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 $= 1 . 3 \mu V \approx 1 \mu V$