The current sensitivity of a moving coil galvanometer can be increased by:
A. Decreasing the magnetic field
B. Increasing the number of turns
C. Increasing the area of the coil
D. Inserting a soft iron core in the coil
E. Decreasing the area of the coil
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → B, C, D Only
The current sensitivity of a moving coil galvanometer is the deflection per unit current. In equilibrium the magnetic torque \( NIAB \) balances the torque \( k\phi \) of the spring, so
\( \displaystyle \frac{\phi}{I} = \frac{NAB}{k} \)
Here N is the number of turns, A the area of the coil, B the magnetic field and k the torsional constant of the spring.
(A) Decreasing the magnetic field — Incorrect. The sensitivity is proportional to B, so it would decrease.
(B) Increasing the number of turns — Correct. The sensitivity is proportional to N.
(C) Increasing the area of the coil — Correct. The sensitivity is proportional to A.
(D) Inserting a soft iron core in the coil — Correct. The soft iron core strengthens the magnetic field in the gap where the coil moves (and makes it radial), so B in the formula is larger.
(E) Decreasing the area of the coil — Incorrect. The sensitivity would decrease.
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 – “… We define the current sensitivity of the galvanometer as the deflection per unit current. From Eq. (4.26) this current sensitivity is,
\( \displaystyle \frac{\phi}{I} = \frac{NAB}{k} \qquad (4.27) \)
A convenient way for the manufacturer to increase the sensitivity is to increase the number of turns N.”