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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