Question

A closely wound solenoid of $2000$ turns and area of cross-section $1.5\times {10}^{–4}{\text{m}}^2$ carries a current of $2.0\text{A}$. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5\times {10}^{–2}\text{tesla}$ making an angle of ${30}^{\circ }$ with the axis of the solenoid. The torque on the solenoid will be

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

Answer 1

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

Given,
Number of turns, $N=2000$,
Area of cross-section, $A=1.5\times {10}^{-4}{\text{m}}^2$,
Current, $I=2\text{A}$,
Magnetic field, $B=5\times {10}^{-2}\text{T}$

Magnetic moment,
$M=NIA=2000\times 2\times 1.5\times {10}^{-4}$
$=0.6{\text{Am}}^2$

Now, the torque on the solenoid,
$\tau =MB\sin {30}^{\circ }$
$=0.6\times 5\times {10}^{-2}\times \frac{1}{2}$
$\tau =1.5\times {10}^{-2}\text{Nm}$

Hence, $\left(C\right)$ is the correct answer.