Show that the current flowing through a moving coil galvanometer is directly proportional to the angle of deflection of coil.

Explanation

Consider a rectangular coil PQRS of length `l` breadth `b` and carrying a current `I` suspended in a uniform magnetic field to induction `B`.
The sides `QR` and `SP` are parallel to `vecB` so that no force acts on them.

The forces `vecF_(1)` and `vecF_(2)` on the sides `PQ` and `RS` are equal in magnitude `(=I//B)`, opposite in direction but have different lines of action, because current flowing in them are in opposite direction,
The magnitude of the deflecting torque due to this couple is
`tau_(d)="Force"xx"Moment arm"`
`=(I//B)b`
`=I(lb)B`
`=IAB`
Where , `A=lb-=` area of the coil. For a coil of `N` turns, practically all in the same plane and having the same area.
`tau_(d)=NIAB`. ..............`(i)`
The torque rotates the coil.
In a moving coil galvanometer, the coil swings in a radial magnetic field produced by the combination of the cylindrically concave pole pieces and the soft iron core.
The rotation of the coil twists the supension fibre which exerts a restoring torque on the coil. The restoring torque is proportional to the angle of twist `theta`.
`tau_(r ) prop theta`
`tau_(r )xxC theta`........`(ii)`
Where `C` is the torque constant , i.e., toorque per unit angle of twist. `C` depends on the dimensions and the elasticity of the supension fibre.
The coil eventually comes to rest in the position where the restoring torque equal to the defelcting torque in magnitude . Therefore, in the equilibrium position,
`tau_(r )=tau_(d)`
`:. C theta=NIAB`.........`(iii)`
`:. I=((C )/(NAB))theta`
`:. I prop theta`
Since , `N`, `A`, `B` and `C` are constants. Thus, the current through the coil is proportional to the deflection of the coil.