Question

Show that the deflection produced in a moving coil galvanometer is directly proportional to the current flowing through its coil or vice versa.

Answer 1

A rectangular coil is suspended in a magnetic field `vecB`.

Let PQRS be a rectangular coil carrying current I.

Let PQ = RS = l = be the length of coil and PS = QR = b = breadth of coil.

The forces acting on PQ and RS are equal in magnitude but opposite in direction. The magnitude of these forces can be computed as shown below.

`F = I  vecl xx vec B`

= I lB sinθ

F = IlB   ..(As θ = 90°, sin90° = 1)

When current I is sent through the coil, these forces acting on PQ and RS form a pair that tends to rotate the coil, i.e. torque, which is given by

τ = Force × Perpendicular distance

= IlB × b

∴ τ = IAB, where, A = l × b = Area of coil 

For N turns of coil,

Torque (τ) = NIAB

Due to this, torque coil deflects. Therefore, it is called deflecting torque.

∴ Deflection torque = NIAB  ...(i)

This pair causes the suspension to twist as the coil rotates. The restoring torque for the coil's deflection (Φ) is given by

Restoring torque = `Kphi`    ...(ii)

where K is the torsional constant of the spring.

In equilibrium condition,

Resting torque = Deflecting torque

K`phi` = NIAB

∴ `phi = ((NAB)/K)I`

∴ `phi ∝ I`, where `(NAB)/K` = constant.

i.e., the deflection produced in a moving coil galvanometer is directly proportional to the current in the coil.