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

Torque exerted on current carrying pivoted rectangular coil when subjected to external magnetic field is,
`tau=BINAsintheta`
(Where `theta` = angle between `vecAandvecB`)
`thereforetau=BIN(l^(2))sintheta`
`thereforetau=(0.8)(12)(20)(0.1)^(2)sin30^(@)" "(becausetheta=30^(@))`
`thereforetau=0.96Nm`