A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 . The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of with the direction of

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Solution

Q.
A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 $Weber / m^{2}$ . The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of $30 °$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be [2015]

1 0.24 N m

2 0.12 N m

3 0.15 N m

4 0.20 N m

Ans.
(4)

The required torque is $τ = N I A B \sin θ$

where $N$ is the number of turns in the coil, $I$ is the current through the coil, $B$ is the uniform magnetic field, $A$ is the area of the coil, and $θ$ is the angle between the direction of the magnetic field and the normal to the plane of the coil.

Here, $N = 50, I = 2 A, A = 0.12 m \times 0.1 m = 0.012 m^{2}$

$B = 0.2$ $W b / m^{2}$ and $θ = 90 ° - 30 ° = 60 °$

$∴ τ = (50) (2 A) (0.012 m^{2}) (0.2 Wb / m^{2})$ $\sin 60 °$ $= 0.20$ $N$ $m$

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