Q.
A solenoid of 0.4m length with 500 turns carries a current of 3A. A coil of 10 turns and of radius 0.01m carries a current of 0.4A. The torque required to hold the coil with its axis at right angles to that of solenoid in the middle point of it is,
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a
6π2×10-7Nm
b
3π2×10-7Nm
c
9π2×10-7Nm
d
12π2×10-7Nm
answer is A.
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Detailed Solution
Field due to solenoid, B=μonI = 4π×10-7×5000.4×3=15π×10-4 T Now, torque experienced by coil in presence of above field Torque =τ=BNiA=BNiπr2=15π×10-4×10×0.4×π×(0.01)2=6π2×10-7Nm
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