A rectangular coil of area \( 40\ \mathrm{cm} \times 20\ \mathrm{cm} \) carrying a current of \( 5\ \mathrm{A} \) is placed in a region of uniform magnetic field of \( 0.2\ \mathrm{T} \) such that its plane makes an angle of \( 30^\circ \) with the direction of the field. If rectangular coil has \( 200 \) turns, then, the torque \( (\mathrm{Nm}) \) acting on the coil is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 8\sqrt{3} \)
The torque on a current loop is \( \tau = NIAB\sin\theta \), where \( \theta \) is the angle between the field and the normal to the plane of the coil.
The plane makes \( 30^\circ \) with the field, so the normal makes \( 90^\circ - 30^\circ = 60^\circ \) with it.
\( A = 40\ \mathrm{cm} \times 20\ \mathrm{cm} = 0.40\ \mathrm{m} \times 0.20\ \mathrm{m} = 0.08\ \mathrm{m^{2}} \)
\( \tau = 200 \times 5\ \mathrm{A} \times 0.08\ \mathrm{m^{2}} \times 0.2\ \mathrm{T} \times \sin60^\circ \)
\( \displaystyle \tau = 16 \times \frac{\sqrt{3}}{2} = 8\sqrt{3}\ \mathrm{N\,m} \)