A copper ring carrying a current is placed in a uniform magnetic field in such a way that the ring experiences maximum possible torque of $$2$$ $$N-m$$. Now the ring is rotated through an angle of $$60o$$ about a diameter which is perpendicular to the field direction. In this position torque experienced by the ring will be

Question

A copper ring carrying a current is placed in a uniform magnetic field in such a way that the ring experiences maximum possible torque of $$2$$ $$N-m$$. Now the ring is rotated through an angle of $$60o$$ about a diameter which is perpendicular to the field direction. In this position torque experienced by the ring will be

Infinity Learn · Accepted Answer

For the position shown in figure, torque experienced by the ring is maximum and $ς_{\max} = M B \sin 90^{o} = M B$ when the ring is rotated through an angle of $60^{o}$ about the diameter PQ, the angle between $\vec{M}$ and $\vec{B}$ is $30^{o}$

$∴ ς = M B \sin 30^{o} = ς_{\max} \times \frac{1}{2} = 2 \times \frac{1}{2} N - m = 1 N - m$

Torque on a current carrying loop placed in uniform magnetic field

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Torque on a current carrying loop placed in uniform magnetic field

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For the position shown in figure, torque experienced by the ring is maximum and ςmax⁡=MBsin⁡90o=MB when the ring is rotated through an angle of 60o about the diameter PQ, the angle between M→ and B→ is 30o∴ς=MBsin⁡30o=ςmax⁡×12=2×12N−m=1N−m

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