When the number of turns is doubled in a moving coil galvanometer, then,
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → current sensitivity is increased but the voltage sensitivity doesn't change.
In a moving coil galvanometer the deflection per unit current (current sensitivity) and the deflection per unit voltage (voltage sensitivity) are
\( \displaystyle I_s = \frac{\phi}{I} = \frac{NAB}{k} \)
\( \displaystyle V_s = \frac{\phi}{V} = \frac{NAB}{kR} \)
Doubling N doubles the current sensitivity. But the coil is wound from a longer wire (twice as many turns of the same wire), so its resistance R also doubles. N and R both double in the voltage sensitivity, so it does not change.
Why the other options are incorrect
Option (1) both current sensitivity and voltage sensitivity are increased. — Incorrect. The voltage sensitivity would increase only if the resistance stayed the same, but more turns means a longer wire and a larger resistance.
Option (2) voltage sensitivity is increased, but current sensitivity is decreased. — Incorrect. Neither part is right: the current sensitivity increases, and the voltage sensitivity stays the same.
Option (4) voltage sensitivity increases but the current sensitivity remains unaffected. — Incorrect. The current sensitivity is directly proportional to N, so it cannot stay unaffected.
NCERT (2026-27):
The Moving Coil Galvanometer – “… An interesting point to note is that increasing the current sensitivity may not necessarily increase the voltage sensitivity. Let us take Eq. (4.27) which provides a measure of current sensitivity. If N → 2N, i.e., we double the number of turns, then
\( \displaystyle \frac{\phi}{I} \rightarrow 2\frac{\phi}{I} \)
Thus, the current sensitivity doubles. However, the resistance of the galvanometer is also likely to double, since it is proportional to the length of the wire. In Eq. (4.28), N → 2N, and R → 2R, thus the voltage sensitivity,
\( \displaystyle \frac{\phi}{V} \rightarrow \frac{\phi}{V} \)
remains unchanged.”