Magnetic induction at the center of a circular loop carrying a current is `'B'` . If `'A'` is the area of the coil, the magnetic dipole moment of the loop is

Options

A. `(BA^(2))/(mu _(0)pi)`
B. `(2BA^(3//2))/(mu_(0)pi ^(1//2))`
C. `(BA^(3//2))/(mu_(0)pi)`
D. `(mu_(0)pi^(1//2))/(BA^(3//2))`

Answer

B

Explanation

`M =1A A =pi R ^(2) therefore R = ((A)/(pi))^(1//2)" "...(1)`
`and B=(mu_(0))/(4pi)((2pi I )/(R))=(mu_(0)I)/(mu_(0))" "...(2)`
`therefore` from (1) and (2)
`therefore M =IA =(2 RB)/(mu_(0))xxA`
`=(2BA)/(mu_(0))xx (A ^(1//2))/(pi ^(1//2))=(2BA ^(3//2))/(mu_(0)pi ^(1//2))`