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A circular coil of $70$ turns and radius $5 c m$ carrying a current of $8 A$ is suspended vertically in a uniform horizontal magnetic field of magnitude $1.5 T$ . The field lines make an angle of $30^{0}$ with the normal of the coil then the magnitude of the counter torque that must be applied to prevent the coil from turning is

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$33 N m$

$3.3 N m$

$3.3 \times 10^{- 4} N m$

$3.3 \times 10^{- 2} N m$

answer is B.

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Detailed Solution

$N = 70, r = 5 c m = 5 \times 10^{- 2} m, I = 8 A B = 1.5 T, θ = 30^{0}$

The counter torque to prevent the coil from turning will be equal and opposite to the torque acting on the coil,

Therefore, $T = N I A B \sin (θ) = N I π r^{2} B \sin (30^{0})$

$= 70 \times 8 \times 3.14 \times {(5 \times 10^{- 2})}^{2} \times 1.5 \times \frac{1}{2} = 3.297 N m = 3.3 N m$

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