Verify the Gauss's law for magnetic field of a point dipole of dipole moment m at the origin for the surface which is a sphere of radius $R$.
Let us draw the figure for given situation,
We have to prove that $\boldsymbol{\delta B} \cdot d \mathbf{S}=0$. This is called Gauss's law in magnetisation.
According to question,
Magnetic moment of dipole at origin $O$ is
$$\mathrm{M}=M \hat{k}$$
Let $P$ be a point at distance $r$ from $O$ and $O P$ makes an angle $\theta$ with $z$-axis. Component of $M$ along $O P=M \cos \theta$.
Now, the magnetic field induction at $P$ due to dipole of moment $\mathrm{M} \cos \theta$ is
$$\mathbf{B}=\frac{\alpha_0}{4 \pi} \frac{2 M \cos \theta}{r^3} \hat{\mathbf{r}}$$
From the diagram, $r$ is the radius of sphere with centre at $O$ lying in $y z$-plane. Take an elementary area $d \mathbf{S}$ of the surface at $P$. Then,
$$\begin{aligned} d \mathbf{S} & =r(r \sin \theta d \theta) \hat{\mathbf{r}}=r^2 \sin \theta d \theta \hat{\mathbf{r}} \\ \oint \cdot d \mathbf{S} & =\oint \frac{\alpha_0}{4 \pi} \frac{2 M \cos \theta}{r^3} \hat{\mathbf{r}}\left(r^2 \sin \theta d \theta \hat{\mathbf{r}}\right) \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{r} \int_0^{2 \pi} 2 \sin \theta \cdot \cos \theta d \theta \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{r} \int_0^{2 \pi} \sin 2 \theta d \theta \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{r}\left(\frac{-\cos 2 \theta}{2}\right)_0^{2 \pi} \\ & =-\frac{\alpha_0}{4 \pi} \frac{M}{2 r}[\cos 4 \pi-\cos 0] \\ & =\frac{\alpha_0}{4 \pi} \frac{M}{2 r}[1-1]=0 \end{aligned}$$
Three identical bar magnets are rivetted together at centre in the same plane as shown in figure. This system is placed at rest in a slowly varying magnetic field. It is found that the system of magnets does not show any motion. The north-south poles of one magnet is shown in the figure. Determine the poles of the remaining two.
The system will be in stable equilibrium if the net force on the system is zero and net torque on the system is also zero. This is possible only when the poles of the remaining two magnets are as given in the figure.
Suppose we want to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole $\mathbf{p}$ in an electrostatic field $\mathbf{E}$ and (ii) magnetic dipole $\mathbf{M}$ in a magnetic field B. Write down a set of conditions on E, B, p, M so that the two motions are verified to be identical. (Assume identical initial conditions).
Now, suppose that the angle between $\mathbf{M}$ and $B$ is $\theta$.
Torque on magnetic dipole moment $\mathbf{M}$ in magnetic field $\mathbf{B}$,
$$\tau^{\prime}=M B \sin \theta$$
$$\begin{aligned} &\text { Two motions will be identical, if }\\ &\begin{aligned} & p E \sin \theta =M B \sin \theta \\ \Rightarrow\quad & p E =M B \\ \text { But, } \quad & E =c B\quad \text{.... (i)} \end{aligned} \end{aligned}$$
$\therefore$ Putting this value in Eq. (i),
$$\begin{aligned} p c B & =M B \\ \Rightarrow\quad p & =\frac{M}{c} \end{aligned}$$
A bar magnet of magnetic moment $M$ and moment of inertia $I$ (about centre, perpendicular to length) is cut into two equal pieces, perpendicular to length. Let $T$ be the period of oscillations of the original magnet about an axis through the mid-point, perpendicular to length, in a magnetic field $B$. What would be the similar period $T^{\prime}$ for each piece?
Given, $I=$ moment of inertia of the bar magnet
$m=$ mass of bar magnet
$l=$ length of magnet about an any passing through its centre and perpendicular to its length
$M=$ magnetic moment of the magnet
$B=$ uniform magnetic field in which magnet is oscillating, we get time period of oscillation is,
$$\begin{aligned} & T=2 \pi \sqrt{\frac{I}{M B}} \\ \text{Here,}\quad & I=\frac{m l^2}{12} \end{aligned}$$
When magnet is cut into two equal pieces, perpendicular to length, then moment of inertia of each piece of magnet about an axis perpendicular to length passing through its centre is
$I^{\prime} =\frac{m}{2} \frac{(l / 2)^2}{12}=\frac{m l^2}{12} \times \frac{1}{8}=\frac{I}{8}$
$$\begin{aligned} \text{Magnetic dipole moment}\quad M^{\prime} & =M / 2 \end{aligned}$$
$$\begin{aligned} \text{ Its time period of oscillation is}\quad T^{\prime} & =2 \pi \sqrt{\frac{I^{\prime}}{M^{\prime} B}}=2 \pi \sqrt{\frac{I / 8}{(M / 2) B}}=\frac{2 \pi}{2} \sqrt{\frac{I}{M B}} \\ T^{\prime} & =\frac{T}{2} \end{aligned}$$
Use (i) the Ampere's law for H and (ii) continuity of lines of $\mathbf{B}$, to conclude that inside a bar magnet, (a) lines of $\mathbf{H}$ run from the N -pole to $S$ - pole, while (b) lines of $\mathbf{B}$ must run from the $S$-pole to $N$-pole.
Consider a magnetic field line of B through the bar magnet as given in the figure below.
The magnetic field line of B through the bar magnet must be a closed loop.
Let $C$ be the amperian loop. Then,
$$\int_Q^P H \cdot d l=\int_Q^P \frac{B}{m_0} \cdot d l$$
We know that the angle between $\mathbf{B}$ and $\mathbf{d l}$ is less than $90^{\circ}$ inside the bar magnet. So, it is positive.
i.e., $$\int_Q^P \mathrm{H} \cdot \mathrm{dl}=\int_Q^P \frac{\mathrm{~B}}{\alpha_0} \cdot \mathrm{dl}>0$$
Hence, the lines of B must run from south pole $(S)$ to north pole $(N)$ inside the bar magnet. According to Ampere's law,
$$\begin{array}{ll} \therefore & \oint_\limits{P Q P} \mathrm{H} . \mathrm{dl}=0 \\ \therefore & \oint_\limits{P Q P} \mathrm{H} \cdot \mathrm{dl}=\int_P^Q \mathrm{H} \cdot \mathrm{dl}+\int_Q^P \mathrm{H} \cdot \mathrm{dl}=0 \\ \text { As } & \int_Q^P \mathrm{H} . \mathrm{dl}>0, \mathrm{so}, \int_P^Q \mathrm{H} . \mathrm{dl}<0 \quad\text{(i.e., negative)} \end{array}$$
It will be so if angle between $\mathbf{H}$ and $\mathbf{d l}$ is more than $90 \Upsilon$, so that $\cos \theta$ is negative. It means the line of H must run from N -pole to S -pole inside the bar magnet.