A rectangular coil of length 0.12 m and width 0.1m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/m2 . 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

A
0.24 Nm
B
0.12 Nm
C
0.15 Nm
D
0.20 Nm

Solution:

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 normal to the plane of the coil. Here, N = 50, I = 2 A, A = 0.12 m × 0.1 m= 0.012 $m^{2}$
B = 0.2 Wb / $m^{2}$ and $θ$ = 90° - 30° = 60°
$∴ τ = (50) (2 A) (0.012 m^{2}) (0.2 Wb / m^{2}) \sin 60 °$

=0.20 Nm

Solution:

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