Question

The workdone in rotating the magnet by ${180}^{\circ }$ from the direction of uniform magnetic field is $W$. The work done in rotating the magnet from the field direction to the position where couple has half of its value when it is at ${90}^{\circ }$ is

• A. $\frac{W}{2}(1-\sqrt{3})$
• B. $2W$
• C. $\frac{\sqrt{3}}{2}W$
• D. $\frac{W}{4}(2-\sqrt{3})$

Answer 1

The correct option is D $\frac{W}{4}(2-\sqrt{3})$

Work done to rotate the magnet is given by, $W=MB[1-\cos \theta ]$

Given , $\theta ={180}^{\circ }\Rightarrow \cos {180}^{\circ }=-1$

$\therefore W=MB[1+1]$

$\Rightarrow W=2MB$ .......(1)

At Half the maximum couple position is $\theta ={30}^{\circ }$

The work done to rotate through ${30}^{\circ }$

$W^{\prime }=MB[1-\cos {30}^o]$

$\Rightarrow W^{\prime }=MB[1-\frac{\sqrt{3}}{2}]$

$\Rightarrow W^{\prime }=\frac{W}{4}(2-\sqrt{3})$

Hence, option (d) is the correct answer.