A long solenoid having 200 turns per cm caries a current of 1.5 amp. At the centre of it 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 sec., the induced e.m.f. 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×10−7200×10−2×1⋅5=3⋅8×10−2Wb/m2Magnetic flux through each turn of the coilΦ=BA=3⋅8×10−23⋅14×10−4=1⋅2×10−5 webreWhen the current in the solenoid is reversed, the change in magnetic flux=2×1⋅2×10−5=2⋅4×10−5 weberInduced e.m.f.=NdΦdt=100×2⋅4×10−50⋅05=0⋅048V