Question

A magnetic dipole lying parallel to a magnetic field requires '\(W\)' units of work to turn it through \(30^{\circ }\). The torque required to keep this needle in this position is
\(\left[\sin 30^{\circ }=\cos 60^{\circ }=0.5,\,\sin 60^{\circ }=\cos 30^{\circ }=\frac{\sqrt{3}}{2},\,\sin 0^{\circ }=\cos 90^{\circ }=1\right]\)

• A. \(\sqrt{3}W\)
• B. \(\frac{2}{\sqrt{3}}W\)
• C. \(\frac{W}{(\sqrt{3}-1)}\)
• D. \(\frac{W}{(2-\sqrt{3})}\)

Hint:

The work done is MB(1 - cos30). Use it to find MB and then the torque MB sin30.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The work in turning a dipole from \(\theta=0\) to \(\theta\) is \(W=MB(1-\cos\theta)\). The torque at angle \(\theta\) is \(\tau=MB\sin\theta\).

Step 2: Find MB
\(W=MB(1-\cos30^{\circ})=MB\left(1-\dfrac{\sqrt{3}}{2}\right)=MB\,\dfrac{2-\sqrt{3}}{2}\)
\(MB=\dfrac{2W}{2-\sqrt{3}}\)

Step 3: Calculation
\(\tau=MB\sin30^{\circ}=\dfrac{2W}{2-\sqrt{3}}\times\dfrac{1}{2}=\dfrac{W}{2-\sqrt{3}}\)

Step 4: Compare the options
This is the last option.

Final Answer:
The torque needed to hold the dipole in that position is W/(2 - root 3). \[ \boxed{\text{(D) }\dfrac{W}{2-\sqrt{3}}} \]