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A square coil of side 10 cm consists of ...

A square coil of side 10 cm consists of 20 turns and carries a current of 12A. The coil is suspended vertically and the normal to the plane makes an angle of `30^(@)` with the direction of uniform magnetic field of 0.8 T. The torque acting on the coil is

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Length of a side of the square coil, 1=10 cm=0.1 m
Current flowing in the coil, I= 12 A
Number of turns on the coil, n=20
Angle made by the plane of the coil with magnetic field, `theta = 30^(@)`
Strength of magnetic field, B=0.80 T
Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,
`tau=nBIAsin theta`
Where, A= Area of the square coil
`=1xx1=0.1xx0.1=0.01 m^(2)`
So,`tau=20xx0.8xx12xx0.01xx sin 30^(@)=0.96Nm`
Hence, the magnitude of the torque experienced by the coil is 0.96 N m.
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