A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the current, the magnitude of the magnetic dipole moment of circular loop will be :

Options

A. `(4m)/(pi)`
B. `(2m)/(pi)`
C. `(3m)/(pi)`
D. `m/(pi)`

Answer

A

Explanation

`m=Ia^(2)` square loop
`m'=Ipir^(2)` circular loop
`4a=2pirimpliesr=(2a)/(pi)` (radius of loop) `implies m'=Ipi((2a)/(pi))^(2)=(4m)/(pi)`