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:
A
6π2×10−7Nm
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B
3π2×10−7Nm
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C
9π2×10−7Nm
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D
12π2×10−7Nm
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Solution
The correct option is A6π2×10−7Nm Magnetic field at the center of solenoid, B=μonI Given: n=5000.4turnsmI=3A B=4π×10−7×5000.4×3=1.5π×10−3T Torque on the coil, τ=NIcoil→A×→B As axis of coil and solenoid are perpendicular to each other, so →A×→B=AB=πr2B Given: r=0.01mN=10Icoil=0.4A τ=10×0.4×π(0.01)2×1.5π×10−3 ⟹τ=6π2×10−7Nm