A coil of area `100cm^(2)` having 500 turns carries a current of `1mA` . It is suspended in a uniform magnetic field of induction `10^(-3)wb//m^(2)` . Its plane makes an angle fo `60^(@)` with the lines of induction. The torque acting on the coil is

Explanation

Given `i= mA = 10^3 = 10^(-3)A, N = 500, B = 10^(-3) Wb//m^2`
`theta = 60^@, tau = ? A =100 cm^2 = 100 xx 10^(-4) m^2`
Couple acting on the coil is given by `tau BiAN sin theta `
Where `phi` is angle made by normal to the plane of coil with B, `phi = 90 - 60 = 30^@`
` therefore C = 10^(-3) xx 10^(-3) xx 100 xx 10^(-4) xx 500 xx sin 30`
` = 250 xx 10^(-8) Nm`