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Torque on a current carrying loop placed in uniform magnetic field

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 $60^{o}$ about a diameter which is perpendicular to the field direction. In this position torque experienced by the ring will be

Moderate

A

1 N-m
B

0.5 N-m
C

1.732 N-m
D

1.414 N-m

Solution

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

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