Question

A point P is situated on the equator of short magnetic dipole of moment of 2 $A{m}^2$. If the distance of that point from its center is 10 cm the magnetic potential at point P will be:

• A. $4\times {10}^{-5}J/Am$
• B. $3.5\times {10}^{-3}J/Am$
• C. $1.55\times {10}^{-6}J/Am$
• D. zero

Answer 1

The correct option is A $4\times {10}^{-5}J/Am$

Given: $M=2A/m^2$

distance of P from centre of dipole,$r=10cm=0.1m$

To find: magnetic potential at P is,$V=?$

Solution: As we know that on axial line,'

$V=\frac{{\mu }_0}{4\pi }\frac{2M}{r^2}$

$==>V={10}^{-7}\frac{2\ast 2}{(0.1{)}^2}$

$==>V=4\ast {10}^{-5}J/Am$