Question

The work done in rotating the magnet from the direction of uniform field to the opposite direction to the field is $W$. The work done in rotating the magnet from the field direction to half the maximum couple position is :

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

Answer 1

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

${\theta }_1=0$
${\theta }_2={180}^o$
$U_1=-mBcos0$
$=-mB$
$U_2=-mBcos180$
$=mB$
From the question,
$W=U_2-U_1$
$W=2mB...........\left(i\right)$
At $\theta ={30}^0$ torque will be half of maximum value as $sin\theta =1/2$

It is required to find work done in rotating the magnet from the field direction to the position of half the torque. Hence,

$W_2=U_3-U_1$
$U_3=-mBcos{30}^0=-mB\frac{\sqrt{3}}{2}$

$W_2=mB(1-\frac{\sqrt{3}}{2})...........\left(ii\right)$

From (i) and (ii),

$W_2=\frac{W}{2}(1-\frac{\sqrt{3}}{2})$

$W_2=\frac{W}{4}(2-\sqrt{3})$