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# 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°with the normal of the coil. Calculate the magnitude of the co

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  $60 \degree$with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.

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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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