A uniform horizontal magnetic field of `7.5xx10^(-2)T` is set up at an angle of `30^(@)` with the axis of an solenoid and the magnetic moment associated with it is `1.28JT^(-1)`. Then the torque on it is

Options

A. `4.8xx10^(-2)Nm`
B. `1.6xx10^(-2)Nm`
C. `1.2xx10^(-2)Nm`
D. `4.8xx10^(-4)Nm`

Answer

A

Explanation

Torque,` tau=MB sin theta`
Here= `1.28JT^(-1), B=7.5xx10^(-2),T=theta=30^(@)`
`therefore tau=1.28xx7.5xx10^(-2)sin 30^(@)`
`=1.28xx7.5xx10^(-2)xx(1)/(2)`
`=0.64xx7.5xx10^(-2)=4.8xx10^(-2)Nm`