Question

# a circular coil of 25 turns and radius 6 cm carrying a current of 10 A, is suspended vertically in a uniform magnetic field of magnitude 1.2 T. The field lines run horizontally in the plane of the coil. What will be the magnitude and dirction of balancing torque required on the coil to prevent it from turning?

A
3.39 N-m, upwards
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B
7.5 N-m, downwards
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C
3.39 N-m, downwards
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D
5 N-m, upwards
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Solution

## The correct option is C 3.39 N-m, downwardsThe torque →τ on a coil of having N turns and current i placed in a magnetic field will be, →τ=−→M×→B |→τ|=NiAB sinθ Since, the coil lies in the plane of →B, θ= 90∘ |→τ|=25×10×[π×(6)2×10−4]×1.2×1 |→τ|=3.39 N-m Using Right-hand thumb rule, the direction of magnetic moment (−→M) is perpendicularly outwards (⊙), Thus using cross product with horizontal magnetic field (rightwards) give the direction for torque as vertically upwards, Thus, to prevent the coil from turning an equal & opposite torque need to be applied. So, the torque must be applied in downward direction. Hence, option (C) is the correct answer.

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