In a moving coil galvanometer with a radial field, the magnetic torque NIAB is balanced by the spring's counter torque kφ. The deflection φ is therefore:

1. In equilibrium the magnetic and restoring torques balance: $k\phi = NIAB$. 2. Here the field is radial, so $\sin\theta = 1$ always. 3. Solve for the deflection: $\phi = \dfrac{NAB}{k}I$. 4. The deflection is directly proportional to the current $I$ (not $I^2$). _Source: NCERT Class 12 Physics Ch 4 "Moving Charges and Magnetism", p.160_