Question

The coil area of a galvanometer is 25 × $10^{-4}m^{2}$. It consists of 150 turns of a wire and is in a magnetic field of 0.15 T. The restoring torque constant of the suspension fibre is $10^{-6}$ Nm per degree. Assuming the magnetic field to be radial, calculate the maximum current that can be measured by the galvanometer, if the scale can accommodate 30° deflection.

Answer 1

Answer: $I = 5.3\times10^{-4}\ A$

Explanation:

Given that,

Area $A = 25\times10^{-4}\ m^{2}$

Magnetic field B = 0.15 T

Torque constant $C = 10^{-6}\ N-m$

Deflection $\theta = 30^{o}$

Number of turns N = 150 turns

We know that,

The maximum current is

$I = \dfrac{C\theta}{NBA}$

$I = \dfrac{10^{-6}\ N-m\times30}{150\times 0.15\ T\times25\times10^{-4}\ m^{2}}$

$I = 5.3\times10^{-4}\ A$

Hence, the maximum current is $5.3\times10^{-4} A$