**Question:**

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

**Solution:**

Length of a side of the square coil, *l* = 10 cm = 0.1 m

Current flowing in the coil, *I* = 12 A

Number of turns on the coil, *n* = 20

Angle made by the plane of the coil with magnetic field, *θ* = 30°

Strength of magnetic field, *B* = 0.80 T

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

τ = *n* *BIA* sin*θ*

Where,

*A* = Area of the square coil

$\Rightarrow \mid \times 1=0.1 \times 0.1=0.01 \mathrm{~m}^{2}$

$\therefore T=20 \times 0.8 \times 12 \times 0.01 \times \sin 30^{\circ}$

= 0.96 N m

Hence, the magnitude of the torque experienced by the coil is 0.96 N m.

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