A long circular cylinder of radius $\mathbf{R}$carries a magnetization $\mathbf{M}=\mathbf{k}{\mathbf{s}}^2\widehat{\varphi }$. Where$\mathbf{k}$is a constant,$\mathbf{s}$is the distance from the axis, and $\widehat{\varphi }$ is the usual azimuthal unit vector (Fig. 6.13). Find the magnetic field due to $\mathbf{M}$, for points inside and outside the cylinder.

Figure 6.13

Consider a long circular cylinder of radius $R$carries a magnetization $\text{M}={\text{ks}}^{\text{2}}\widehat{\varphi }$. Where $k$is a constant, $s$is the distance from the axis, and $\widehat{\varphi }$ is the usual azimuthal unit vector.

Write the formula of magnetic field outside the cylinder.

${\overrightarrow{\mathbf{B}}}_{\mathrm{out}}=\frac{{\mu }_0{I}_{\mathrm{enc}}}{\mathbf{2}\pi \mathbf{s}}$ …… (1)

Here,${\mu }_0$ is permeability, $I_{\mathrm{enc}}$is current enclosed the cylinder and $s$ is the distance from the axis.

Write the formula of magnetic field inside the cylinder.

${\overrightarrow{\mathbf{B}}}_{\mathrm{ins}}=\frac{{\mathit{\mu }}_0}{\mathbf{2}\pi \mathbf{s}}{\int }_S{\mathbf{J}}_b\mathbf{dS}$ …… (2)

Here, ${\mathit{\mu }}_0$ is permeability, $\mathbf{s}$is the distance from the axis and ${\mathit{J}}_{\mathbf{b}}$is surface bound current.

First we determine the bound currents:

$\begin{array}{c}{\overrightarrow{K}}_b=\overrightarrow{M}\times \widehat{s}\\ =-k{R}^2\widehat{z}\end{array}$

Determine the surface bound currents:

$\begin{array}{c}{\overrightarrow{J}}_b=\nabla \times \overrightarrow{M}\\ =\frac{1}{s}\frac{\partial }{\partial s}\left(sk{s}^2\right)\widehat{z}\\ =3ks\widehat{z}\end{array}$

Now, determine the magnetic field outside the cylinder. Use a circular Amperian loop:

$\begin{array}{c}{\overrightarrow{B}}_{out}=2\pi k_b+{\int }_0^R{\int }_0^{2\pi }{J}_{b}sdsd\varphi \\ =\left(2\pi R\right)(-k{R}^2)+3k{\int }_0^R{\int }_0^{2\pi }{s}^{2}ds\\ {\overrightarrow{B}}_{out}=0\end{array}$

Determine the magnetic field inside the cylinder. Use a circular Amperian loop (this time of radius $s<R$):

$\begin{array}{c}{\overrightarrow{B}}_{ins}=\frac{{\mu }_0}{2\pi s}{\int }_S{J}_{b}dS\\ ={\mu }_{0}3k{\int }_0^s{\int }_0^{2\pi }{s}^{2}dsd\varphi \\ ={\mu }_{0}2\pi k{s}^2\\ {\overrightarrow{B}}_{ins}={\mu }_{0}k{s}^2\widehat{\varphi }\end{array}$

Therefore, the value of magnetic field outside and inside the cylinder is${\overrightarrow{B}}_{out}=0$ and ${\overrightarrow{B}}_{ins}={\mu }_{0}k{s}^2\widehat{\varphi }$.