Q. 16 B 4.8( 4 Votes )

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.

Calculate the induced emf in the loop at the instant when x = 0.2 m.

Take a = 0.1 m and assume that the loop has a large resistance.

Answer :

The emf induced in the coil is given as:

e = B’ av

e =

substituting the given values in above relation, we get

e =

On calculating, we get

e = 5 × 10-5V

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