Analysis of Magnetized Sphere Statements
This solution analyzes the properties of an isolated magnetized sphere with uniform magnetization $\vec{M}$ along the positive $z$ direction.
Bound Volume Current Density ($\vec{J}_b$)
The bound volume current density is calculated using $\vec{J}_b = \nabla \times \vec{M}$.
Given $\vec{M} = M\hat{z}$ is uniform:
$ \vec{J}_b = \nabla \times (M\hat{z}) = 0 $
Since $\vec{J}_b = 0$, the first statement is CORRECT.
Bound Surface Current Density ($\vec{K}_b$)
The bound surface current density is $\vec{K}_b = \vec{M} \times \hat{n}$ on the surface.
Here, $\hat{n} = \hat{r}$ (outward normal) and $\vec{M} = M\hat{z}$. Using spherical coordinates ($\hat{z} = \cos\theta \hat{r} - \sin\theta \hat{\theta}$):
$ \vec{K}_b = (M\hat{z}) \times \hat{r} = M(\cos\theta \hat{r} - \sin\theta \hat{\theta}) \times \hat{r} = M\sin\theta \hat{\phi} $
The magnitude $|\vec{K}_b| = M |\sin\theta|$. It is maximum ($M$) at the equator ($\theta = \pi/2$).
Therefore, the second statement is CORRECT.
Auxiliary Field ($\vec{H}$) Calculation
The relation is $\vec{B} = \mu_0 (\vec{H} + \vec{M})$.
Inside the sphere, $\vec{B}_{in} = \frac{2\mu_0}{3} \vec{M}$. Substituting this:
$ \frac{2\mu_0}{3} \vec{M} = \mu_0 (\vec{H}_{in} + \vec{M}) $
Solving for $\vec{H}_{in}$:
$ \vec{H}_{in} = \frac{2}{3}\vec{M} - \vec{M} = -\frac{1}{3}\vec{M} $
The third statement claims $\vec{H} = -\frac{2}{3}\vec{M}$, which is incorrect.
Therefore, the third statement is INCORRECT.
Far-Field Magnetic Field Behavior
Far from the sphere ($r \gg R$), the magnetic field resembles that of a dipole $\vec{m}$.
The dipole moment for a uniformly magnetized sphere is $\vec{m} = \frac{4}{3}\pi R^3 \vec{M}$.
Let's check the relation given in the fourth statement: $\frac{\vec{m}}{4\pi R^3} = \frac{B}{2\mu_0} \hat{z}$. We use the internal field $B = B_{in} = \frac{2\mu_0}{3} M$.
LHS: $\frac{\vec{m}}{4\pi R^3} = \frac{1}{4\pi R^3} (\frac{4}{3}\pi R^3 M \hat{z}) = \frac{1}{3} M \hat{z}$.
RHS: $\frac{B}{2\mu_0} \hat{z} = \frac{1}{2\mu_0} (\frac{2\mu_0}{3} M \hat{z}) = \frac{1}{3} M \hat{z}$.
Since LHS = RHS, the fourth statement is CORRECT.
Correct Statements Summary
The analysis shows that statements A, B, and D are correct.
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Column I
Column II 1. 
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(R) Ferromagnetic 4. 
(S) Antiferromagnetic
