The magnetic potential at a point on the line inclined at `30^(@)` with the axis of a short magnet is `1.5xx10^(-5) J//Am`If the magnetic moment is `1.732 Am^(2)`,What will be distance of point from centre of the magnet?

Options

A. 10cm
B. 1 cm
C. 0.1 cm
D. 0.001cm

Answer

B

Explanation

`V=(mu_(0))/(4pi) (Mcostheta)/(V)`
`therefore r^(2)=(mu_(0))/(4pi)(Mcostheta)/(V)`
`r^(2)=(mu_(0))/(4pi)(Mcostheta)/(V)`
`r^(2)=(4pixx10^(-7)xx1.732xxcos30)/(4pixx1.5xx10^(-5))`
`=`0.9999xx10^(-2)`
`therefore r=0.09999m 10 cm`