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. $\frac{{\mu }_{0}M}{4\pi z^3}$
• B. $\frac{2{\mu }_{0}M}{4\pi z^3}$
• C. $\frac{4\pi M}{{\mu }_0{z}^3}$
• D. $\frac{{\mu }_{0}M}{2\pi z^3}$

Answer 1

The correct option is B $\frac{2{\mu }_{0}M}{4\pi z^3}$

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
Similarly on S-Pole
$B_2=\frac{{\mu }_0}{4\pi }\frac{M}{(z+l{)}^2}$ along SP

Net magnetic flux at P is
$B=B_1-B_2=\frac{{\mu }_0}{4\pi }\left[\frac{M}{(z-l{)}^2}-\frac{M}{(z+l{)}^2}\right]$
$=\frac{{\mu }_0}{4\pi }\left[\frac{4Mzl}{(z^2-l^2{)}^2}\right]$
$=\frac{{\mu }_0}{4\pi }\left[\frac{M\times 2l}{(z^2-l^2{)}^2}2z\right]$

$\therefore B=\frac{{\mu }_0}{4\pi }\frac{2Mz}{(z^2-l^2{)}^2}$

$B=\frac{{\mu }_0}{4\pi }\cdot \frac{2M}{z^3}$ $(\because z>>l^2)$