Torque on a current carrying loop placed in uniform magnetic field

Question

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/${m}^{2}$. 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

Difficult

Solution

The required torque is

$\tau =NIAB\mathrm{sin}\theta $

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 $\theta $ 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 $\theta $ = 90° - 30° = 60°

$\therefore \tau =\left(50\right)(2\mathrm{A})\left(0.012{\mathrm{m}}^{2}\right)\left(0.2\mathrm{Wb}/{\mathrm{m}}^{2}\right)\mathrm{sin}60\xb0$

=0.20 Nm

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