A moving coil galvanometer has a circular coil of area `16 cm^(2)` and 30 turns. It is mounted on a torsional spring of spring constant 100 Nm/rad. The coil turns by angle `21^(@)` when a current 40 mA is passed through it. The magnitude of the uniform magnetic field produced by the permanent magnet inside the galvanometer is _________________T.
[use `cos 21^(@)=15/16, 21^(@)~~11/30` rad]

Answer

A

Explanation

The galvanometer coil rotates in a uniform magnetic field because of a torque given by:
`vec(tau)=vecmxxvecB`
Here, `vecm` is the magnetic moment of the coil and `vecB` is the magnetic field.
The construction of the galvanometer is such that initially the magnetic moment vector is perpendicular to the magnetic field, and the torsional spring is in a relaxed state. As the torque acts on the coil, the rotation of the coil leads to a deformation of the spring. An equilibrium is achieved where the torque due to the magnetic field is balanced by the restoring torque due to the spring.
Let the angular displacement of the coil from the original position be . Then, the restoring torque due to the spring is given by:`tau_(S)=C theta` , Here C is the torsional constant of the spring.
In this condition, the torque due to the magnetic field is,
`|vec(tau)_(B)|=|vecmxxvecB|=m B sin (90^(@)-theta)=m B cos theta`
Since in equilibrium `tau_(B)=tau_(S)`
`m B cos theta =C thetaimpliesB=(C theta)/(m cos theta)=((5.4xx10^(-3))(11/30))/((0.04)(16xx10^(-4))(30)(15/16))=1.1T`