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Home » A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniformthe horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform
the horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
CBSE | Class 12 | Moving Charges and Magnetism | NCERT | Physics
A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30º with the direction of a uniform
the horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

The length of a square coil’s side (l) is 0.1 m.

The current flowing through the coil (I) has a magnitude of 12 A.

The number of coil turns (n) is 20.

The angle formed by the coil’s plane with B (Magnetic field), \theta = {30^ \circ }

The magnetic field strength (B) is 0.8 T.

The magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the following relationship:

\tau = nBIA\sin \theta

Where,

A = Area of the square coil

= 0.1 x 0.1 = 0.01 {m^2}

So,

    \[\tau = 20 \times 0.8 \times 12 \times 0.01 \times \sin {30^ \circ }\]

= 0.96 N m

As a result, the magnitude of the torque experienced by the coil is 0.96 N m.

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