A coil in the shape of an equilateral triangle of side 0.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of `5 xx 10^(-2) T.` Find the couple acting on the coil when a current of 0.1 A is passed through it and the magnetic field is parallel to its plane.

Explanation

As the coil is in the form of an equilateral triangle, its area
`A=(1)/(2)xxLxxLsin60^(@)`
`=(1)/(2)xx(0.02)^(2)xx(sqrt(3))/(2)=sqrt(3)xx10^(-4)m^(2)`

So its magnetic moment
`M=iA=0.1xxsqrt(3)xx10^(-4)`
`=sqrt(3)xx10^(-5)A-m^(2)`
Now, the couple on a current-carrying coil in a magnetic field is given by `tau=MBsintheta`
Since the plane of the coil is parallel to the magnetic field, the angle between M and `vecB` will be `90^(@)` and hence `tau=MBsin90^(@)=MB`
`therefore tau=(sqrt(3)xx10^(-5))xx5xx10^(-2)=5sqrt(3)xx10^(-7)N-m`