A long solenoid having 200 turns per centimeter carries a current of 1.5 A. At the center of the solenoid is placed a coil of 100 turns of cross-sectional area 3.14 x 10-4 m2 having its axis parallel to the field produced by the solenoid. When the direction of current in the solenoid is reversed within 0.05 s, the induced emf in the coil is
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a
0.48 V
b
0.048 V
c
0.0048 V
d
48 V
answer is B.
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Detailed Solution
B=μ0ni=4π×10−7×200×102×1.5=3.8×10−2Tϕ=BA=3.8×10−2×3.14×10−4=1.2×10−5WbWhen the current in the solenoid is reversed, the change in magnetic flux,dϕ=2×1.2×10−5=2.4×10−5Wb∴ Induced e.m.f. ε=Ndϕdt=100×2.4×10−50.05=0.048V