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.

Options

A. `0.866xx10^(-6)N-M`
B. `1.732xx10^(-4)N-M`
C. `0.422xx10^(-6)N-M`
D. `0.866xx10^(-2)N-M`

Answer

1

Explanation

Here `l=0.02m, B=5xx10^(-2)T,I=0.1A`
Area of the triangular coil,
`A=(1)/(2)|vec(l)xxvec(l)|=(1)/(2)(lxxlsing60^(@))`
`=(1)/(2) l^(2) sin 60^(@)=(1)/(2)(0.02m)^(2)(0.865)=1.732xx10^(-4)m^(2)`
`(` as angle between the sides of an equilateral triagle is `60^(@))`
Couple acting on the coil,`,tau = IAB cos alpha`
`=(0.1A)(1.32xx10^(-4)m^(2))(5xx10^(-2)T)cos 0^(@)`
`=0.866xx10^(-6)Nm`