Question

A closely wound solenoid of 2000 turns and area of cross-section $1.5\times {10}^{-4}{m}^2$ 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 $5\times {10}^{-2}$ tesla making an angle of 30${}^o$ with the axis of the solenoid. The torque on the solenoid will be

• A. $3\times {10}^{-3}Nm$
• B. $1.5\times {10}^{-3}Nm$
• C. $1.5\times {10}^{-2}Nm$
• D. $3\times {10}^{-2}Nm$

Answer 1

The correct option is C $1.5\times {10}^{-2}Nm$

Magnetic moment of the loop.
M = NIA = $2000\times 2\times 1.5\times {10}^{-4}=0.6$ J/T
torque $\tau$ = MBsin30${}^o$
$=0.6\times 5\times {10}^{-2}\times \frac{1}{2}$
$=1.5\times {10}^{-2}Nm$