To find the deflection of a moving coil galvanometer, the torque on i…

The correct answer is Option (3) → the field is radial, so the plane of the coil always lies along the field

For a coil of \( N \) turns the torque is \( NIAB\sin\theta \), where \( \theta \) is the angle between the normal to the coil and the field. The torque becomes \( NIAB \) only when \( \theta=90^\circ \).

Option (1) the coil turns only through small angles, so \(\sin\theta\) stays close to 1 — Incorrect. The size of the deflection is not the reason. In a uniform field the angle \( \theta \) changes as the coil turns. Also, a small \( \theta \) makes \( \sin\theta \) small, not close to 1.

Option (2) the soft iron core makes the field uniform, so the torque does not change with angle — Incorrect. The core makes the field radial and stronger. In a uniform field the torque would depend on the position of the coil.

Option (3) the field is radial, so the plane of the coil always lies along the field — Correct. In a radial field the field lines lie in the plane of the coil in every position. So the normal to the coil is always at \( 90^\circ \) to the field, and \( \sin\theta=1 \).

Option (4) the field is radial, so it always points along the normal to the coil — Incorrect. Radial field lines lie in the plane of the coil. If the field were along the normal, \( \theta \) would be zero and so would the torque.

NCERT (2026-27):
The Moving Coil Galvanometer – “There is a cylindrical soft iron core which not only makes the field radial but also increases the strength of the magnetic field.”

The Moving Coil Galvanometer – “… Since the field is radial by design, we have taken sin θ = 1 in the above expression for the torque.”