Question

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s^-1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Answer 1

Length of the rectangular wire, l = 8 cm = 0.08 m 
Width of the rectangular wire, b = 2 cm = 0.02 m 
Hence, area of the rectangular loop,
A = lb
= 0.08 x 0.02
= 16 x 10^-4 m²
Magnetic field strength, B = 0.3 T Velocity of the loop, v = 1 cm/s = 0.01 m/s 
(a) Emf developed in the loop is given as: 
e = Blv
= 0.3 × 0.08 x 0.01 = 2.4 x 10^-4 V
Time taken to travel along the width, t = \(\frac{Distance travelled}{Velocity}\) = \(\frac{b}{v}\)

                                                                = \(\frac{0.02}{0.01}\)= 2 s
Hence, the induced voltage is 2.4 x 10^-4 V which lasts for 2 s.

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(b) Emf developed, e = Bbv
= 0.3 × 0.02 x 0.01 = 0.6 x 10^-4 V
Time taken to travel along the length, t = \(\frac{Distance travelled}{Velocity}\) = \(\frac{I}{V}\)
                                                                = \(\frac{0.08}{0.01}\) = 8 s
Hence, the induced voltage is 0.6 x 10^-4 V which lasts for 8 s.