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Physics - Magnetism Question with Solution | TestHub

PhysicsMagnetismMagnetic torque & energyMedium2 minPYQ_2016

PhysicsMediumsingle choice

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 isNEET 2016 (Phase 2)

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Answer:

Solution:

$W = P E (\cos θ_{1} - \cos θ_{2})$

$W = P E (\cos 0 - \cos 60^{o})$

$= \frac{P E}{2}$ .

$\Rightarrow P E = 2 W$

$τ = P E \sin θ = 2 W \sin 60^{o} = \sqrt{3} W$ .

Stream:NEETSubject:PhysicsTopic:MagnetismSubtopic:Magnetic torque & energy

⏱ 2mℹ️ Source: PYQ_2016

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