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
$B=B_1-B_2=\frac{{\mu }_0}{4\pi }\left[\frac{M}{(z-l{)}^2}-\frac{M}{(z+l{)}^2}\right]$
$=\frac{{\mu }_0}{4\pi }\left[\frac{4Mdl}{(z^2-l^2{)}^2}\right]$
$=\frac{{\mu }_0}{4\pi }\left[\frac{M\times 2l}{(z^2-l^2{)}^2}2z\right]$
$B=\frac{{\mu }_0}{4\pi }\frac{2Mz}{(z^2-l^2{)}^2}$
$B=\frac{{\mu }_0}{4\pi }.\frac{3Md}{z^2}(\because z>>l^2)$
$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2M}{z^3}\hat{M}$