Looking for classes? Ksquare Career Institute, Bengaluru →
A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position is:
A
$\frac{\sqrt{3}W}{2}$
B
$\frac{2W}{\sqrt{3}}$
C
$\frac{W}{\sqrt{3}}$
D
$\sqrt{3}W$
Explanation
Divide $MB\sin\theta$ by $MB(1-\cos\theta)$.
Detailed Solution
$\tau = MB\sin60^\circ$ ...(1)
$W = MB(1 - \cos60^\circ)$ ...(2)
From (1) and (2): $\frac{\tau}{W} = \frac{\sqrt{3}/2}{1/2} \Rightarrow \tau = \sqrt{3}W$
$W = MB(1 - \cos60^\circ)$ ...(2)
From (1) and (2): $\frac{\tau}{W} = \frac{\sqrt{3}/2}{1/2} \Rightarrow \tau = \sqrt{3}W$