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Figure shows one quarter of a single circular loop of wire that carries a current of 14 A. Its radius is 5 cm. A uniform magnetic field B = 300 G is directed along +x direction. The torque acting on the loop is

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$2 .9 \times 10^{- 3} Nm$

$1 .9 \times 10^{- 3} Nm$

6.2 Nm

$0.5 \times 10^{- 3} Nm$

answer is A.

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Detailed Solution

The normal to the loop, OP makes an angle $θ = 60^{0}$ with +x direction, i.e., along the field direction.

$\begin{array}{l} Hence, Torque τ = NiAB \sin θ \\ τ = 1 \times 14 \times π (25 \times 10^{- 4}) (0 .03) \times \sin 60^{0} \\ = 2 .9 \times 10^{- 3} N - m \end{array}$

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