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  • Now assume that the straight wire carries a current of 50 A and the loop is moved to the right with a constant velocity v=10 m/ s.

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.

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best_answer

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.

Posted by

Pankaj Sanodiya

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