Question

The magnetic field due to short bar magnet of magnetic dipole moment $M $ and length $2l$, on the axis at a distance $z$ (where $z >> l$) from the centre of the magnet is given by formula

• A. $\frac{\mu_0 M}{4\pi z^3}M$
• B. $\frac{2\mu_0 M}{4\pi z^3}M$
• C. $\frac{4\mu_0 M}{\mu_0 z^3}M$
• D. $\frac{4\mu_0 M}{2\pi z^3}M$

Answer 1

Answer: (B)

Consider a point $P$ located on the axial line of a short bar magnet of magnetic length $2l$ and strength $M$.
Let us find $B$ at a point $P$ which is at a distance $z$ from the centre of magnet.
Magnetic flux density of $P$ due to $N$-pole is
$B_1=\frac{\mu_0}{4\pi}\left(\frac{M}{(z-l)^2}\right)$ along $NP$
Similarily on $S$-Pole
$B_2=\frac{\mu_0}{4\pi}\left(\frac{M}{(z-l)^2}\right)$ along $SP$
Net magnetic flux at Pis
$B=B_1-B_2=\frac{\mu_0}{4\pi}\left[\frac{M}{(z-l)^2}-\frac{M}{(z+l)^2}\right]$
$=\frac{\mu_0}{4\pi}\left[\frac{4Mdl}{(z^2-l^2)^2}\right]$
$=\frac{\mu_0}{4\pi}\left[\frac{M\times 2l}{(z^2-l^2)^2}2z\right]$
$B=\frac{\mu_0}{4\pi}\frac{2Mz}{(z^2-l^2)^2}$
$B=\frac{\mu_0}{4\pi}.\frac{3Md}{z^2}\,\, (\because z > > l^2)$
$\therefore B=\frac{\mu_0}{4\pi} \frac{2M}{z^3}\hat{M}$