A moving coil galvanometer has a coil of effective area A and number of turns N. The magnetic field B is radial. If a current I is passed through the coil, the torque acting on the coil is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( NABI \)
A coil of \( N \) turns carrying current \( I \) has magnetic moment \( m = NIA \). In a field \( B \) the torque on it is
\( \displaystyle \tau = mB\sin\theta = NIAB\sin\theta \)
In a radial magnetic field the plane of the coil always lies along the field lines, so the angle between \( \vec{m} \) and \( \vec{B} \) is always \( 90^\circ \) and \( \sin\theta = 1 \).
\( \displaystyle \tau = NABI \)
The torque is the same in every position of the coil, which makes the deflection proportional to the current.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
The Moving Coil Galvanometer – “… The galvanometer consists of a coil, with many turns, free to rotate about a fixed axis …, in a uniform radial magnetic field. … When a current flows through the coil, a torque acts on it. This torque is given by Eq. (4.20) to be
\( \displaystyle \tau = NIAB \)
where the symbols have their usual meaning. Since the field is radial by design, we have taken sin θ = 1 in the above expression for the torque.”