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Q.

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. The magnitude of torque experienced by the coil is

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a

1.36 N-m

b

0.96 N-m

c

2.06 N-m

d

0.23 N-m

answer is A.

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Detailed Solution

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Given, side of square coil =10 cm = 0.1 m
Number of turns (n) = 20
Current in square coil I =12 A
Angle made by coil q = 30°
Magnetic field B = 0.80 T
The magnitude of torque experienced by the coil
$\begin{array}{l} τ = N I A B \sin θ \\ = 20 \times 12 \times {(10 \times 10^{- 2})}^{2} \times 0.80 \times \sin 30^{0} \\ τ = 2.4 \times 0.80 \sin 30^{0} = \frac{2.4 \times 0.80}{2} = 0.96 N - m \end{array}$

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