JEE Main 2023 — Electromagnetic Induction Question with Solution

Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.

This statement is true. The formula for current sensitivity ($I_s$) of a moving coil galvanometer is given by:

$I_s=\frac{NAB}{k}$

where:

• $N$ is the number of turns in the coil,
• $A$ is the area of the coil,
• $B$ is the magnetic field strength, and
• $k$ is the spring constant of the coil.

From this formula, you can see that the current sensitivity is directly proportional to the number of turns (N). If $N$ is doubled, then the current sensitivity will also double.

Statement II: Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also increase its voltage sensitivity in the same ratio.

This statement is false. The formula for voltage sensitivity ($V_s$) of a moving coil galvanometer is given by:

$V_s=I_{s}R=\frac{NAB}{k}R$

where:

• $R$ is the resistance of the coil.

From this formula, you can see that the voltage sensitivity is proportional to the number of turns ($N$) but also inversely proportional to the coil resistance ($R$). If you double the number of turns ($N$), you also double the length of the wire making up the coil, and thus, you double the resistance ($R$) of the coil. The doubling of $N$ is offset by the doubling of $R$, so the overall voltage sensitivity remains the same.

Therefore, increasing the current sensitivity by only increasing the number of turns in the coil will not increase the voltage sensitivity in the same ratio.