A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is ranging in a vertical in plane magnetic field of 5 x 10^–2 T. Find the couple acting on the coil when a current of 0.1 ampere is passed through it and the magnetic field is parallel to its plane.

Solution

Given,
Side of equilateral triangle, a = 0.02 m
Magnetic field, B = 5 x 10^–2 T
Current passd through the coil, I = 0.1 A

As the coil is in the form of an equilateral triangle, its area is

A = \frac{\sqrt{3}}{4} \times 0.02 \times 0.02 A = \sqrt{3} \times 10^{- 4} m = \sqrt{3} \times 10^{- 4} m^{2}

Torque on current carring coil in magnetic field is

τ = BINA sinα

where,

α = 90 °

N = 1

τ = BIA

= 5 \times 10^{- 2} \times 0.1 \times \sqrt{3} \times 10^{- 4} N^{- m} = 0.5 \times \sqrt{3} \times 10^{- 6} = 5 \sqrt{3} \times 10^{- 7} N m

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