Q6.16 (b) Now assume that the straight wire carries a current of $50A$ and the loop is moved to the right with a constant velocity, $V=10m/s$. Calculate the induced emf in the loop at the instant when $X=0.2m$. Take $a=0.1m$ and assume that the loop has a large resistance.

Given,

Current in the straight wire

$I$= 50A

Speed of the Loop which is moving in the right direction

$V=10m/s$

Length of the square loop

$a=0.1m$

distance from the wire to the left side of the square

$X=0.2m$

NOW,

Induced emf in the loop :

$E=B_xav=\frac{\mu _0I}{2\pi x}av=\left ( \frac{4\pi*10^{-7}*50}{2\pi*0.2} \right )*0.1*10=5*10^{-5}V$

Hence emf induced is $5*10^{-5}V$.

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