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

\frac{\sqrt{3} W}{2}

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\frac{2 W}{\sqrt{3}}

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\frac{W}{\sqrt{3}}

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\sqrt{3} W

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Solution

The correct option is D $\sqrt{3} W$

Torque acting on the magnet is, $τ = M B sin 60 °_____(1) Work done can be written as, W = M B (1 - cos 60 °)(2) From (1) and (2) \frac{τ}{W} = \frac{\sqrt{3} / 2}{1 / 2} \Rightarrow τ = W \sqrt{3}$

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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

The correct option is D √3W Torque acting on the magnet is,τ = MB sin 60° _________ (1)Work done can be written as,W = MB (1 − cos 60°) ____(2)From (1) and (2)τW = √3/21/2⇒τ = W√3

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

The correct option is D $\sqrt{3} W$

Torque acting on the magnet is, $τ = M B sin 60 °_____(1) Work done can be written as, W = M B (1 - cos 60 °)(2) From (1) and (2) \frac{τ}{W} = \frac{\sqrt{3} / 2}{1 / 2} \Rightarrow τ = W \sqrt{3}$