A square shaped plane coil of area `100cm^2` of `200` turns carries a steady current of `5A`. It is placed in a uniform magnetic field of `0*2T` acting perpendicular to the plane of the coil. Calculate the torque on the coil when its plane makes an angle of `60^@` with the direction of the field. In which orientation will the coil be in stable equilibrium?

Explanation

Here, `A=100cm^2=100xx10^-4m^2`
`=10^-2m^2`,
`n=200, I=5A, B=0*2T, theta=90^@-60^@=30^@`
Torque on coil, `tau=nIABsintheta`
`=200xx5xx10^-2xx0*2xxsin30^@=1Nm`
The coil will be in a stable equilibrium when the resultant force and torque on the coil will be zero. The torque will be zero when `theta=0^@`, i.e., the plane of the coil is perpendicular to the direction of magnetic field.