A vertical coil of sides `5cm xx 2cm` has 10 turns and carries `2A` current. Calculate torque on the coil, when it is placed in a uniform horizontal magnetic field of `0*1T` with its plane at `60^@` to the field.

Answer

`10^-3N-m`

Explanation

Here, `A=5xx2=10cm^2=10xx10^-4m^2`
`=10^-3m^2`
`N=10`, `I=2A`, `tau=?` `B=0*1T`
`theta=90^@-60^@=30^@`
`M=NIA=10xx2xx10^-3=2xx10^-2Am^2`
`tau=MBsintheta`
`=2xx10^-2xx0*1sin30^@=10^-3N-m`