Magnetic Field Intensity at a Point on the Axial Line of a Bar Magnet - Magnetism and Matter - Param Himalaya

Let O be the centre of a bar magnet having magnetic length 2l. Let p be the point on the axial line of the bar magnet at a distance r from the centre O of the bar magnet.

Magnetic Field Intensity at a Point on the Axial Line of a Bar Magnet - Magnetism and Matter - Param Himalaya

Let $q_{m}$ be the pole strength of each pole of the magnet. Let a unit north pole be placed at point P. 

Assume that the presence of unit N-pole at P does not affect the magnetic field at P due to the bar magnet.

Magnetic field intensity at P due to north pole of the bar magnet, $\overrightarrow{B_{1}}$ = Force experienced by unit north pole at P due to north pole of the magnet ,

$$\overrightarrow{B_{1}} = \frac{F}{q_{0}}= \frac{\frac{\mu_{0}}{4\pi}\frac{q_{m}q_{0}}{(r-l)^{2}}}{q_{0}}$$

$$\overrightarrow{B_{1}} = \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r-l)^{2}} along NP$$

$$\overrightarrow{B_{1}} = \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r-l)^{2}}\hat{i}$$

similarly , magnetic field intensity at a point P due to south pole of the  bar magnet ,

$$\overrightarrow{B_{2}} = \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r+l)^{2}}\hat{-i}$$

$\therefore$ Net magnetic field intensity at a point P due to the bar magnet.

$$\overrightarrow{B_{a}} = \overrightarrow{B_{1}}+\overrightarrow{B_{2}}$$

$$\overrightarrow{B_{a}}= \left [ \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r-l)^{2})} - \frac{\mu_{0}}{4 \pi}\frac{q_{m}}{(r+l)^{2})}\right  ]\hat{i}$$

$$\overrightarrow{B_{a}} = \frac{\mu_{0}}{4\pi}q_{m}\frac{(r+l)^{2}-(r-l)^{2}}{(r^{2}-l^{2})^{2}}\hat{i}$$

$$\overrightarrow{B_{a}} = \frac{\mu_{0}}{4\pi}q_{m}\left [ \frac{2r\times 2l}{(r^{2}-l^{2})} \right ]\widehat{i}$$

since  $q_{m} \times 2l=m$ , dipole moment

$$\overrightarrow{B_{a}} = \frac{\mu_{0}}{4\pi}\left [ \frac{2mr}{(r^{2}-l^{2})} \right ]\widehat{i}$$

The magnitude of the magnetic field due to a bar magnet at a distance r on th axial line of the bar magnet from the centre of the bar magnet is given by 

$$\overrightarrow{B_{a}} = \frac{\mu_{0}}{4\pi}\left [ \frac{2mr}{(r^{2}-l^{2})} \right ]$$

if the magnet is of very small length , then $l^{2} <<r^{2}$

$$\therefore \overrightarrow{B_{a}} = \frac{\mu_{0}}{4\pi}\left [ \frac{2mr}{(r^{3}} \right ]\widehat{i}$$

Direction of $\overrightarrow{B_{a}}$ is along the direction  of magnetic dipole moment $\overrightarrow{m}$. Thus , angle between $\overrightarrow{B_{a}}$ and $\overrightarrow{m}$





































Previous Post Next Post