Question

Establish the formual of intensity of magnetic field due to a bar magent in the equatorial position.

Answer 1

Lets NS be a bar magnet of pole strength m and effective length $2l$. Consider a point P on the neutral axis at a distance d from the centre of the magnet.
Let $PN=PS=x$
Now, the intensity of field at P due to N-pole is,
$B_1=\frac{{\mu }_o}{4\pi }\cdot \frac{m}{x^2}$(along $\overrightarrow{NP}$)
Similarly, the intensity of field at P due to S pole is given by $B_2=\frac{{\mu }_o}{4\pi }\cdot \frac{m}{x^2}$(along $\overrightarrow{PS}$)
Since, $B_1=B_2=\frac{{\mu }_o}{4\pi }\cdot \frac{m}{x^2}$
Hence,
Hence, by the law of parallelogram of force the resultant will be given by
$B=\sqrt{B_1^2+B_2^2+2B_1{B}_{2}cos2\theta }$
(where $2Q$ is the angle between $B_1$ and $B_2$)
But, $B_1=B_2$, hence
$B=\sqrt{B_1^2+B_1^2+2B_1^{2}cos3\theta }$
$=\sqrt{2B_1^2+2B_1^{2}cos2\theta }$
$=\sqrt{2B_1^2(1+cos2\theta )}$
$=\sqrt{2B_1^2[1+2co{s}^2\theta -1]}$
$=\sqrt{2B_1^2\cdot 2co{s}^2\theta }$
$=2B_{1}cos\theta$
or $B=2\cdot \frac{{\mu }_o}{4\pi }\cdot \frac{m}{x^2}\cdot cos\theta$
or $B=\frac{{\mu }_o}{4\pi }\cdot \frac{2m}{x^2}\cdot \frac{1}{x},(\because cos\theta =\frac{1}{x})$
or $B=\frac{{\mu }_o}{4\pi }\cdot \frac{2ml}{x^3}$
But, $2ml=M$(magnetic moment)
$\therefore B=\frac{{\mu }_o}{4\pi }\cdot \frac{2M}{x^3}$
But, by the Pythagoras theorem in right angle $\Delta$PQS.
$x=\sqrt{d^2+l^2}$
or $x^3=(d^2+l^2{)}^{3/2}$
Hence, $B=\frac{{\mu }_o}{4\pi }\cdot \frac{M}{(d^2+l^2{)}^{3/2}}$
If the magnet is small i.e., $d>>1$
$thereforeB=\frac{{\mu }_o}{4\pi }\cdot \frac{M}{d^3}$
This is the required expression.
The direction of magnetic field is from north pole to south pole and parallel to magnetic axis of the magnet.