A square loop of area 25 cm 2 has a resistance of 10 Ω . The loop is placed in uniform magnetic field of magnitude 40 . 0 T . The plane of loop is perpendicular to the magnetic field. The work done in

Question

A square loop of area 25 cm 2 has a resistance of 10 Ω . The loop is placed in uniform magnetic field of magnitude 40 . 0 T . The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1 . 0 sec, will be

TestHub · Accepted Answer

As l 2 = 25 cm 2 , then l = 5 cm = 0 . 05 m . Given: t = 1 s Velocity of the square loop, v = 0 . 05 1 = 0 . 05 m s - 1 . Induced current, i = V R = B l v R = 40 × 0 . 05 × 0 . 05 10 = 0 . 01 A Now force acting on the side of the square loop, F = B i l = 40 × 0 . 01 × 0 . 05 ⇒ F = 0 . 02 N Therefore, work done W = F l = 0 . 02 × l = 0 . 02 × 0 . 05 ⇒ W = 1 × 10 - 3 J

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