Due to the flow of current in a circular loop of radius `R`, the magnetic induction produced at the centre of the loop is `B`. The magnetic moment of the loop is (`mu_(0)`=permeability constant)

Options

A. `BR^(3)//2pimu_(0)`
B. `2piBR^(3)//mu_(0)`
C. `BR^(2)//2pimu_(0)`
D. `2piBR^(2)//mu_(0)`

Answer

B

Explanation

`B=(mu_(0)i)/(2R)impliesi=(Bxx2R)/(mu_(0))`
Now, `M=ixxA=ipiR^(2)=(Bxx2R)/(mu_(0))xxpiR^(2)=(2piBR^(3))/(mu_(0))`