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.

Answer

`8*66xx10^-7N-m`

Explanation

Let the coil of triangular area A carrying current I be suspended in a uniform magnetic field `B(=5xx10^-2T)`. Let `theta` be the angle between magnetic field and normal to the surface of the loop.
Here, area, `A=1/2("base" x x"height")`
base`=0*02m`,
height`=0*02sin60^@=0*07132m`.
Then torque acting on the closed loop is
`tau=BIA sin theta`
`=(5xx10^-2)xx0*1xx1/2xx(0*02xx0*07132)xxsin90^@`
`=8*66xx10^-7Nm`