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Magnetism and Matter — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsMagnetism and Matter5 Marks
Question
Explain end-on position and broad side-on position of a bar magnet.
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Answer

(a) End-on-position of a bar magnet : Let NS be a bar magnet of length $2l$ and 'm' be the pole strength each of its magnetic poles. Suppose P is a point along the axis of the bar magnet at a distance r from the middle point O of the magnet at which the magnetic field intensity due to the bar magnet is to be determined.
This position is called a end on position or longitudinal position. The distance of point P from north pole N is equal to $(r-l)$ and from south pole S is equal to $(r+l)$ [Fig.(a)].
Image
The magnitude of the magnetic field intensity vec $\overrightarrow{ B _1}$ due to north pole N is given by:
$B_1=\frac{\mu_0}{4 \pi}\left[\frac{m}{(r-l)^2}\right]$ ...(1)
Direction of $\overrightarrow{ B _1}$ is away from the magnet along the axis.
The magnitude of the magnetic field intensity $\overrightarrow{ B _2}$ due to south pole S is given by :
$B_2=\frac{\mu_0}{4 \pi}\left[\frac{m}{(r+l)^2}\right]$ ...(2)
Direction of $\overrightarrow{ B _2}$ is towards the magnet along the axis.
From equations (1) and (2) it is obvious that $B _1> B _2$ and $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ both are oppositely directed. Hence the magnitude of the resultant magnetic field $\overrightarrow{ B }$ at P is given by:
$B = B _1- B _2$
$=\frac{\mu_0}{4 \pi}\left[\frac{m}{(r-l)^2}\right]-\frac{\mu_0}{4 \pi}\left[\frac{m}{(r+l)^2}\right]$
$=\frac{\mu_0 m}{4 \pi}\left[\frac{1}{(r-l)^2}-\frac{1}{(r+l)^2}\right]$
or $B =\frac{\mu_0 m}{4 \pi}\left[\frac{(r+l)^2-(r-l)^2}{(r-l)^2(r+l)^2}\right]$
$=\frac{\mu_0 m}{4 \pi}\left[\frac{2 \times 2 r l}{\{(r-l)(r+l)\}^2}\right]$
or $B =\frac{\mu_0}{4 \pi}\left[\frac{2 \times(m \times 2 l) \times r}{\left(r^2-l^2\right)^2}\right]$
But $m \times 2l$ = Magnetic dipole moment of the bar magnet = M
$\therefore \quad B =\frac{\mu_0}{4 \pi}\left[\frac{2 Mr }{\left(r^2-l^2\right)^2}\right]$ ...(3)
If distance r >> $l$ the magnet is called short bar magnet and in this case in equation (3) $l^2$ can be neglected in comparison to r². Hence magnetic field intensity for a short bar magnet is given by :
$B=\frac{\mu_0}{4 \pi}\left[\frac{2 Mr}{\left(r^2\right)^2}\right] \Rightarrow B =\frac{ \mu _0}{ 4 \pi}\left[\frac{ 2 M }{ r ^3}\right] \ldots$(4)
(b) Broad side on position : In this position the point of observation lies on the perpendicular bisector of the bar magnet. This position is called as broad side on position or transverse position.
Let in Fig. (b), P be a point in the broad side on position of the bar magnet NS of length $2l$ at a distance $r$ from its centre O and let $m$ be the pole strength of each of its magnetic poles. From Fig. 8.10 it is evident that the distance of the point P from each of the poles N and S is equal and is given as follows :
$NP = SP =\left(r^2+ l ^2\right)^{1 / 2}$
Image
The magnitude of magnetic field intensity $\overrightarrow{ B _1}$ at P due to north pole N is given by :
$B _1=\frac{\mu_0}{4 \pi}\left(\frac{m}{ NP ^2}\right)$
$\Rightarrow \quad B _1=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(\sqrt{\left.r^2+l^2\right)^2}\right.}\right]$
$\Rightarrow \quad B_1=\frac{\mu_0}{4 \pi}\left[\frac{ m }{\left( r ^2+ l ^2\right)}\right]$... (1)
The direction of $\overrightarrow{ B _1}$ at P is away from the north pole N.
The magnitude of magnetic field intensity $\overrightarrow{ B _2}$ at P due to south pole S is given by :
$B _2=\frac{\mu_0}{4 \pi}\left(\frac{m}{ SP ^2}\right)$
$\Rightarrow \quad B _1=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(\sqrt{\left(r^2+l^2\right)}\right)^2}\right]$
$\Rightarrow \quad B _2=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(r^2+l^2\right)}\right]$ ...(2)
The direction of vec $\overrightarrow{ B _2}$ at P is towards the south pole S. From equations (1) and (2) it is obvious that :
$B _1= B _2=\frac{\mu_0}{4 \pi}\left[\frac{m}{\left(r^2+l^2\right)}\right]$
In order to find out the resultant magnetic field at P due to the bar magnet; we have to resolve $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ into two rectangular components: one along equitorial line and the other along axial line.
If $\angle PSO =\theta$, then $\angle SPX ^{\prime}=\theta$ and $\angle ONP =\angle X ^{\prime} PA ^{\prime}$ $=\theta$
The components of vec $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ along equatorial line are $B _1 \sin \theta$ and $B _2 \sin \theta$ respectively. As $B _1= B _2$, hence these components will be equal in magnitude, but are opposite in direction. Therefore these will cancel each other. The components of $\overrightarrow{ B _1}$ and $\overrightarrow{ B _2}$ along axial line will be $B _1 \cos \theta$ and $B _2 \cos \theta$ respectively and these will be added up; because these are in the same direction.
Therefore the resultant magnetic field at P due to bar magnet is given by :
$B = B _1 \cos \theta+ B _2 \cos \theta$
[Along $PX ^{\prime}$ ]
$\Rightarrow \quad B =\frac{\mu_0}{4 \pi}\left[\left(\frac{m}{\left(r^2+l^2\right)}\right) \cos \theta\right]$ $+\frac{\mu_0}{4 \pi}\left[\left(\frac{m}{r^2+l^2}\right) \cos \theta\right]$
or $B =\frac{\mu_0}{4 \pi}\left(\frac{2 m}{r^2+l^2}\right) \cos \theta$
But from right angled triangle $\Delta SOP$, we have
$\cos \theta=\frac{ SO }{ SP }=\frac{l}{\left(r^2+l^2\right)^{1 / 2}}$
$\therefore \quad B =\frac{\mu_0}{4 \pi}\left[\frac{2 m}{\left(r^2+l^2\right)}\right] \times \frac{l}{\left(r^2+l^2\right)^{1 / 2}}$
$\Rightarrow \quad B =\frac{\mu_0}{4 \pi}\left[\frac{2 m l}{\left(r^2+l^2\right)^{3 / 2}}\right]$
$\Rightarrow \quad B =\frac{\mu_0}{4 \pi}\left[\frac{m \times 2 l}{\left(r^2+l^2\right)^{3 / 2}}\right]$
But $m \times 2 l=$ Magnetic dipole moment of the bar magnet = M.
$\therefore \quad B =\frac{\mu_0}{4 \pi}\left[\frac{ M }{\left(r^2+I^2\right)^{3 / 2}}\right]$ ...(3)
Direction of $\overrightarrow{ B }$ is along $PX ^{\prime}$ i.e. from north pole towards south pole parallel to the magnetic axis. Hence its direction is opposite to that of the magnetic dipole moment of the magnet which is directed from south pole towards north pole.
Hence equation in its vector form can be written as follows :
$\vec{B}=-\frac{\mu_0}{4 \pi}\left[\frac{\vec{M}}{\left(r^2+l^2\right)^{3 / 2}}\right]$ ... (4)
For a short bar magnet i.e. a bar magnet of very small length we have $l \ll r$, hence in equations (3) and (4) $l^2$ can be neglected in comparison to $r^2$. Therefore these equations can be written in the following form respectively.
$B=\frac{\mu_0}{4 \pi}\left(\frac{M}{r^3}\right)$ ... (5)
$\overrightarrow{B}=-\frac{\mu_0}{4 \pi}\left(\frac{\overrightarrow{M}}{r^3}\right)$ ... (6)

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