91Ó°ÊÓ

Consider a short circular cylinder of radius and length carries a "frozen-in" uniform magnetization parallel to its axis.

Write the formula of bound current.

${\stackrel{\mathbf{→}}{\mathbf{K}}}_{\mathbf{b}}\mathbf{=}\stackrel{\mathbf{→}}{\mathbf{M}}\mathbf{×}\stackrel{\mathbf{→}}{\mathbf{s}}..........\left(1\right)$

Here, $\stackrel{\mathbf{→}}{\mathbf{M}}$is frozen-in uniform magnetization and$\stackrel{\mathbf{→}}{\mathbf{s}}$ is distance from axis.

Draw the circuit diagram of magnetic field of the cylinder for L>>a.

Figure 1

Draw the circuit diagram of magnetic field of the cylinder for L<<a.

Figure 2

Draw the circuit diagram of magnetic field of the cylinder for $L≈a$.

Figure 3

Determine the bound current is given by:

Substitute $\widehat{\mathrm{ϕ}}$for $\stackrel{→}{s}$into equation (1).

${\stackrel{→}{k}}_b=M\widehat{\mathrm{ϕ}}$

Thus, the field is made up of a solenoid with the following dimensions: L, a, and for the surface current.

The lines of the magnetic field are drawn above. They resemble the bar electret case in appearance, but inside they are entirely different since the magnetic field does not have discontinuities at the top or bottom like the electric field does.