Question

A magnetic needle lying parallel to a magnetic field requires W units of work to turn it through${60}^{\circ }$ . The torque required to maintain the need in this position will be

• A. $\sqrt{3}W$
• B. $W$
• C. $e^{\frac{\sqrt{3}}{2}}W$
• D. $2W$

Answer 1

Answer: (A)

$\sqrt{3}W$

$\begin{array}{c}\text{Let the torgue (t) on the needle due to}\\ \text{magnetic field. be-}\\ \begin{array}{cc}\tau =mB\sin \theta & m\to \text{magnetic moment}\\ B\to \text{magnetic field}& \end{array}\\ \begin{array}{c}\theta \to \text{angle between}\\ \text{magnetic moment and}\\ \text{magnetic field.}\end{array}\end{array}$

$\begin{array}{c}\Rightarrow \text{work done in Rotating it from}{0}^{\circ }⟶{60}^{\circ }\\ \Rightarrow \omega ={\int }_0^{\pi /3}\overrightarrow{\tau }\cdot d\theta ={\int }_0^{\pi /3}mB\sin \theta \cdot d\theta \\ \omega =\frac{mB}{2}-0\end{array}$

$\begin{array}{c}\text{torque at}\theta ={60}^{\circ }=\frac{\pi }{3}\text{is}\\ \begin{array}{c}\tau =mB\frac{\sqrt{3}}{2}-\left(2\right)\\ \text{From eq}\left(1\right)\text{and}\left(2\right)\text{we get}\\ \tau =\sqrt{3}\omega \end{array}\end{array}$