A 300 turns rectangular coil of length 20 cm and breadth 12 cm is carrying a current of 12 A, when placed in a magnetic field of 6 T. The plane of the coil is making an angle of 60° with the magnetic field. What is the torque acting on the coil?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 259.2 N m
A current loop has a magnetic moment \( \vec{m} = N I \vec{A} \), directed along the normal to the plane of the coil. In a uniform field it feels a torque \( \vec{\tau} = \vec{m}\times\vec{B} \), of magnitude
\( \tau = N I A B \sin\theta \)
Here \( \theta \) is the angle between \( \vec{B} \) and the normal. The plane of the coil makes 60° with the field, so the normal makes
\( \theta = 90^\circ - 60^\circ = 30^\circ \)
\( A = 0.20\,\mathrm{m} \times 0.12\,\mathrm{m} = 0.024\,\mathrm{m^2} \)
\( \displaystyle \tau = 300 \times 12 \times 0.024 \times 6 \times \sin 30^\circ = 518.4 \times 0.5 = 259.2\,\mathrm{N\,m} \)