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 $5\times {10}^{-2}T$, making an angle of 30${}^o$ with the axis of the solenoid. The torque on the solenoid will be

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

Answer 1

Answer: (C)

$1.5\times {10}^{-2}N-m$

Given : $N=2000$ turns $A=1.5\times {10}^{-4}{m}^2$ $I=2.0$ A $B=5\times {10}^{-2}$ T $\theta ={30}^o$

Torque $\tau =NI(A\times B)=NIAB\sin \theta$

$\therefore$ $\tau =\left(2000\right)\left(2.0\right)(1.5\times {10}^{-4})(5\times {10}^{-2})\left(0.5\right)$ $(\because \sin {30}^o=0.5)$

$⟹$ $\tau =1.5\times {10}^{-2}$ $N-m$