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A rectangular coil of wire of 500 turns of area 50 cm² carries a current of 2A in a magnetic field of induction 20 gauss. If the torque on the coil is 5 × 10⁴ dynecm, angle between $(B)$ and normal to plane of coil is

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30°

45°

60°

90°

answer is A.

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Detailed Solution

$\begin{array}{l} t o r q u e = niABsin θ = 5 \times 10^{4} \times 10^{- 7} \\ = 500 \times 2 \times 50 \times 10^{- 4} \times 20 \times 10^{- 4} \times \sin θ \\ θ = 30^{0} \end{array}$
$θ$ ’ = 30⁰ ; This is angle between field and plane

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