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 `30^@` with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Explanation

We are given : `l = 10 cm = (10)/(100 ) m = 0.1 m `
Area of coil, `A = 0. 1 xx 0.1 = 10 ^(-2) m ^(2)`
`N = 20 , l = 1 2 A, theta = 30^(@), B = 0.80 T`
We are to calculate `(tau)` torque
Using relation :` tau =NBlA = sin theta = 20 xx 0.80 xx 12 xx 10 ^(-2) sin 30^(@) = 1. 92 xx (1)/(2) = 0.96 N m.`