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The Electrostatic Analog

The Electrostatic Analog :- In electrostatics we see the following results due to an electric dipole at a point located far away from it :-

Electric field on the axis of an electric dipole, $\displaystyle {{E}_{axial}}=\frac{2kp}{{{r}^{3}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{2p}{{{r}^{3}}}$ …..(1)

Electric field at the equator (equatorial line) of an electric dipole, $\displaystyle {{E}_{equatorial}}=\frac{kp}{{{r}^{3}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{p}{{{r}^{3}}}$ …..(2)

Torque experienced by an electric dipole in an external electric field, $\displaystyle \overrightarrow{\tau }=\overrightarrow{p}\times \overrightarrow{E}$ …..(3)

Potential energy of an electric dipole in an external electric field, $\displaystyle U=-\overrightarrow{p}.\overrightarrow{E}$ …..(4)

Now if we make the following substitutions in the above equations :-

$\displaystyle \overrightarrow{p}\to \overrightarrow{M}$

$\displaystyle \overrightarrow{E}\to \overrightarrow{B}$

$\displaystyle \frac{1}{4\pi {{\varepsilon }_{0}}}\to \frac{{{\mu }_{0}}}{4\pi }$

Then we will get similar results for a magnetic dipole (bar magnet) with magnetic dipole moment M and length 2l :-

Magnetic field on the axis of a bar magnet (if r >> l), $\displaystyle {{B}_{axial}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2M}{{{r}^{3}}}$ …..(5)

Magnetic field at the equator of a bar magnet (if r >> l), $\displaystyle {{B}_{equatorial}}=\frac{{{\mu }_{0}}}{4\pi }\frac{M}{{{r}^{3}}}$ …..(6)

The torque experienced by a bar magnet in the external magnetic field, $\displaystyle \overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$ …..(7)