A moving coil galvanometer has N number of turns in the coil with area A. It carries a current I and the magnetic field is B. The torque acting on the coil is :
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( NIBA \)
A current loop in a magnetic field has a magnetic moment \( m = NIA \) and feels a torque \( \tau = mB\sin\theta \), where \( \theta \) is the angle between the normal to the coil and the field.
In a moving coil galvanometer the field is radial, so the plane of the coil always contains the field lines and \( \theta = 90^\circ \).
\( \displaystyle \tau = NIA\,B\sin90^\circ = NIBA \)
The torque is proportional to the first power of each of \( N \), \( I \), \( B \) and \( A \), so the squared forms in the other options cannot be right.
NCERT (2026-27):
The Moving Coil Galvanometer – “… When a current flows through the coil, a torque acts on it. This torque is given by Eq. (4.20) to be
\( \displaystyle \tau = NI AB \)
where the symbols have their usual meaning. Since the field is radial by design, we have taken \( \sin\theta = 1 \) in the above expression for the torque. The magnetic torque NIAB tends to rotate the coil.”