The dipole moment of a coil of 200 turns carrying a current of `3A` is `9*24Am^2`. What is the diameter of the coil?

Answer

`14cm`

Explanation

Here, `N=200`, `I=3A`, `M=9*24Am^2d=?`
If r is radius of coil, then
`M=NIA=Nipi(d^2)/(4)`
`d=sqrt((4M)/(piNI))=sqrt((4xx9*24)/(3*14xx200xx3))`
`=0*14m=14cm`