A current I flows in a conducting wire of lenth L. If we bent it in a circular form, then calculate its magnetic dipole moment.

Options

A. `(lL^(2))/(4 pi)Am^(2) `
B. `(l^2L)/(4 pi ) Am^(2)`
C. `(lL^(2))/(2 pi ) Am^(2) `
D. `(l^(2) L)/(2 pi )Am^(2)`

Answer

A

Explanation

Let a wire of length L is bent in a circular form of radius r.
Then , ` 2pi r=Limplies r=(L)/(2pi)` .......... (i)

The magnetic dipole moment of a circular ring, `M=lA` ( where , A is area of the ring )
or `M=lpir^(2)` ............ (ii)
On putting the value of r from Eq. (i) in Eq. (ii) , we get
`M=lpi ((L)/( 2pi))^(2)`
`implies M=lpi xx (L^2)/( 4 pi^(2))`
` implies M=(lL^(2))/(4pi)A-m^(2)`