The potential at a point at a distance of 25 cm from the centre of a magnetic dipole on a line inclined at angle at `60^(@)` with its dipole moment vector is greater than the potential at a point on the same line at a distance of 50 cm from the centre,by `3xx10^(-7) J//Am`.`The magnetic moment of the magnet is

Options

A. `0.1 Am^(2)`
B. `2.5 Am^(2)`
C. `1.5 Am^(2)`
D. `0.5 Am^(2)`

Answer

D

Explanation

`V_(1)-V_(2)=(mu_(0))/(4pi)M[(1)/(r_(1)^(2))-(1)/(r_(2)^(2))]`
`3xx10^(-7)=10^(-7)xxM[(1)/(6.25xx10^(2))-(1)/(25xx10^(-2))]`
`therefore M=0.5 Am^(2)`