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 $60^{o}$ is $W$ , Now the torque required to keep the magnet in this new position is.

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

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

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\sqrt{3} W

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

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Solution

The correct option is A $\sqrt{3} W$
Torque acting on a bar magnet kept in a magnetic field $\to B$ is $τ = \to m \times \to B = m B s i n θ$

Hence work done in moving from $θ = 0^{\circ}$ to $θ = 60^{\circ}$ is $W = m B c o s 60^{\circ} = \frac{1}{2} m B$

Hence torque acting when magnet is placed at $60^{\circ} = \frac{\sqrt{3}}{2} m B = \sqrt{3} W$

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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 60o is W, Now the torque required to keep the magnet in this new position is.

The correct option is A √3WTorque acting on a bar magnet kept in a magnetic field →B is τ=→m×→B=mBsinθHence work done in moving from θ=0∘ to θ=60∘ is W=mBcos60∘=12mBHence torque acting when magnet is placed at 60∘=√32mB=√3W

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 $60^{o}$ is $W$ , Now the torque required to keep the magnet in this new position is.