4.13) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of  60 \degreewith the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

Answers (1)
S Sayak

Number of turns in the coil(n)=30

The radius of the circular coil(r)=8.0 cm

Current flowing through the coil=6.0 A

Strength of magnetic field=1.0 T

The angle between the field lines and the normal of the coil=60o

The magnitude of the counter-torque that must be applied to prevent the coil from turning would be equal to the magnitude of the torque acting on the coil due to the magnetic field.

\\\tau =nBIAsin\theta\\ \tau = 30\times 1\times 6\times \pi \times (0.08)^{2}\times sin60^{o}

=3.13 Nm

A torque of magnitude 3.13 Nm must be applied to prevent the coil from turning.

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