Question

A moving coil galvanometer has been fitted with a rectangular coil having 50 turns and dimensions 5 cm × 3 cm. The radial magnetic field in which the coil is suspended is of 0.05 Wb/m². The torsional constant of the spring is 1.5 × 10^-9 Nm/ degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of 30°.

Answer 1

Data : N = 50, C = 1.5 × 10^-9 Nm/degree,
A = lb = 5 cm × 3 cm = 15 cm² = 15 × 10^-4 m²,
B = 0.05Wb/m², θ = 30°
NIAB = Cθ
∴ The current through the coil, I = \(\frac{C \theta}{N A B}\)
= \(\frac{1.5 \times 10^{-9} \times 30}{50 \times 15 \times 10^{-4} \times 0.05}=\frac{3 \times 10^{-5}}{5 \times 0 \cdot 5}\)
= 1.2 × 10^-5 A