Verify the Ampere’s law for magnetic field of a point dipole of dipole moment ${\rm{\vec M = M\hat k}}$. Take $\mathrm{C}$ as the closed curve running clockwise along : the $\mathrm{z}$ - axis from $\mathrm{z} = \mathrm{a} \,>\, 0$ to $\mathrm{z = R}$;
From $P$ to $Q$, every point on the $z$-axis lies at the axial line of magnetic dipole of moment $M$. Due to this magnetic moment the point $(0,0, \mathrm{Z})$ at $z$ distance the magnetic field induction,
$\mathrm{B}=2\left(\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{z^{3}}\right)$
$\mathrm{B}=\frac{\mu_{0} \mathrm{M}}{2 \pi z^{3}}$
$(i)$ From Ampere's law, At point from $\mathrm{P}$ to $\mathrm{Q}$ along $z$-axis
$\int_{\mathrm{P}}^{\mathrm{Q}} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l} =\int_{\mathrm{P}}^{\mathrm{Q}} \mathrm{B} d l \cos 0^{\circ}=\int_{a}^{\mathrm{R}} \mathrm{B} d z$
$=\int_{a}^{\mathrm{R}} \frac{\mu_{0}}{2 \pi} \frac{\mathrm{M}}{z^{3}} d z=\frac{\mu_{0} \mathrm{M}}{2 \pi}\left(-\frac{1}{2}\right)\left(\frac{1}{\mathrm{R}^{2}}-\frac{1}{a^{2}}\right)$
$=\frac{\mu_{0} \mathrm{M}}{4 \pi}\left(\frac{1}{a^{2}}-\frac{1}{\mathrm{R}^{2}}\right)$
A short bar magnet has a magnetic moment of $0.48\; J \;T ^{-1} .$ Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10 \,cm$ from the centre of the magnet on
$(a)$ the axis,
$(b)$ the equatorial lines (normal bisector) of the magnet.
Magnetic moments of two bar magnets may be compared with the help of
A short magnet is allowed to fall along the axis of a horizontal metallic ring. Starting from rest, the distance fallen by the magnet in one second may be.....$m$
A bar magnet is placed north-south with its north pole due north. The points of zero magnetic field will be in which direction from the centre of the magnet
Following figures show the arrangement of bar magnets in different configurations. Each magnet has magnetic dipole moment $\vec m$ . Which configuration has highest net magnetic dipole moment