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 $30^{\circ}$ 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. $1.5 \times 10^{-2} Nm$
• D. $3 \times 10^{-2} Nm$

Answer 1

Answer: (C)

Magnetic moment of the loop.
$M=N I A=2000 \times 2 \times 1.5 \times 10^{-4}=0.6 J / T$
Torque $\tau= MB \sin ^{\circ} 30^{\circ}$
$=0.6 \times 5 \times 10^{-2} \times \frac{1}{2}=1.5 \times 10^{-2} N m$