A closely wound solenoid of `200` turns and area of cross-section `1.5xx10^(-4)m^(2)` carries a current of `2.0 A` It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field `5xx10^(-2)T`, making an angle of `30^(@)` with the axis of the solenoid. The torque on the solenoid will be

Options

A. `3xx10^(-3)N-m`
B. `1.5xx10^(-3)N-m`
C. `1.5xx10^(-2)N-m`
D. `3xx10^(-2)N-m`

Answer

c

Explanation

`tau=NiBA sin theta`
`=2000xx2xx5xx10^(-2)xx1.5xx10^(-4)xxsin theta`
`= 2000xx50xx10^(-6)xx(1)/(2)= 1.5xx10^(-2)N-m`