A rectangular coil of wire of 50 turns and dimensions 6cm×9cm carries a current of 10mA. If a magnetic field of magnitude 0.5T is applied parallel to the plane of the coil. What is the magnitude of the torque acting on the coil? Also find the magnitude of the torque when the magnetic field makes an angle of 30degree and 0degree with the normal to the plane of the coil.
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Answer:
Number of turns on the circular coil, n = 100
Radius of each turn, r = 8.0 cm = 0.08 m
Current flowing in the coil, I = 0.4 A
Magnitude of the magnetic field at the centre of the coil is given by the relation,
| B | = μ0 2πnI / 4π r
Where,
μ0 = Permeability of free space
= 4π × 10–7 T m A–1
| B | = 4π x 10-7 x 2π x 100 x 0.4 / 4π x 0.08
= 3.14 x 10-4 T
Hence, the magnitude of the magnetic field is 3.14 × 10–4 T
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