Moving Coil Galvanometer — Working Principle

Question

A rectangular coil of 50 turns, area $2 \times 10^{-4}$ m², carrying a current of 1 mA, is placed in a uniform magnetic field of 0.1 T. The spring constant of the suspension is $2 \times 10^{-6}$ N·m/rad. Find the deflection of the galvanometer and its current sensitivity.

Solution — Step by Step

Why This Works

The key design feature of a moving coil galvanometer is the radial magnetic field created by a cylindrical soft iron core placed inside the coil. Because the field is always parallel to the plane of the coil at every angle, the torque $\tau = nBIA\sin\theta$ reduces to $\tau = nBIA$ throughout the rotation. Without this, the relationship between current and deflection would be non-linear and calibration would be impossible.

The phosphor bronze suspension wire provides the restoring torque. A softer wire (smaller $k$) gives more deflection for the same current — that’s why high-sensitivity galvanometers use very fine suspension fibres. This trade-off between sensitivity and ruggedness is a real design constraint, and NEET/JEE love asking questions about what happens to sensitivity when you change $n$, $B$, $A$, or $k$.

Alternative Method — Using Voltage Sensitivity

If the question gives terminal voltage $V$ across the galvanometer instead of current:

Since $I = V/R_g$, the voltage sensitivity is:

Note the inverse dependence on $R_g$. Increasing current sensitivity by increasing $n$ does NOT necessarily increase voltage sensitivity — more turns means higher resistance $R_g$, so the two effects partially cancel. This specific insight appeared in JEE Main 2024 Shift 1.

Common Mistake