A circular coil of 20 turns and radius 10 cm is placed in a uniform magnetic field of 0.10 T normal to the plane of the coil. If the current in the coil is 5 A. The total torque on the coil is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Zero
The torque on a current loop in a uniform field is \( \vec{\tau} = N\,\vec{m}\times\vec{B} \), that is \( \tau = NIAB\sin\theta \), where \( \theta \) is the angle between \( \vec{B} \) and the normal (area vector) of the coil.
The field is normal to the plane of the coil, so \( \vec{B} \) is parallel to the area vector, \( \theta = 0^\circ \).
\( \tau = NIAB\sin0^\circ = 0 \)
The data (20 turns, 10 cm, 5 A) are not needed; the torque is zero whatever their values.