Read the following passage and answer the questions.
A school laboratory has a moving coil galvanometer. Its rectangular coil has 100 turns, and each turn has an area of 4.0 cm2. The coil can turn between the curved pole pieces of a magnet, with a soft iron cylinder inside it. The magnetic field at the coil is 0.25 T. A spring gives a restoring torque of \(2.0\times10^{-5}\ \mathrm{N\,m}\) for each radian of twist. The coil has a resistance of 50 Ω. The pointer reaches the end of the scale for a current of 2 mA. A student first uses the galvanometer as a current detector in a Wheatstone bridge. Later she plans to change it into an ammeter and into a voltmeter, so that she can measure larger currents and voltages in other circuits.
The current sensitivity of the galvanometer (deflection per unit current) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(500\ \mathrm{rad\,A^{-1}}\)
In equilibrium, the spring torque \( k\phi \) balances the magnetic torque \( NIAB \) on the coil. So the deflection per unit current is \( \phi/I=NAB/k \).
Given:
\( N=100 \)
\( A=4.0\ \mathrm{cm^{2}}=4.0\times10^{-4}\ \mathrm{m^{2}} \)
\( B=0.25\ \mathrm{T} \)
\( k=2.0\times10^{-5}\ \mathrm{N\,m\,rad^{-1}} \)
\( \begin{aligned} \frac{\phi}{I} &= \frac{NAB}{k} \\ &= \frac{100\times4.0\times10^{-4}\times0.25}{2.0\times10^{-5}} \\ &= \frac{10^{-2}}{2.0\times10^{-5}} \\ &= 500\ \mathrm{rad\,A^{-1}} \end{aligned} \)
Therefore, a current of 1 mA turns the coil through 0.5 rad.
NCERT (2026-27):
The Moving Coil Galvanometer – “… We define the current sensitivity of the galvanometer as the deflection per unit current. … this current sensitivity is,”
\( \displaystyle \frac{\phi}{I}=\frac{NAB}{k} \)\( \qquad (4.27) \)