Question

A short circular cylinder of radius and length L carries a "frozen-in" uniform magnetization M parallel to its axis. Find the bound current, and sketch the magnetic field of the cylinder. (Make three sketches: one forL>>a, one forL<<a, and one for$L\approx a$.) Compare this bar magnet with the bar electret of Prob. 4.11.

Short answer

The value of bound current is ${\overrightarrow{K}}_b=M\hat{\varphi }$.

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

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

Draw the magnetic field of the cylinder for $L\approx a$.

Step-by-step solution

Write the given data from the question.

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

Determine the formula of bound current.

Write the formula of bound current.

${\overrightarrow{\mathbf{K}}}_{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{M}}\mathbf{\times }\overrightarrow{\mathbf{s}}..........\left(1\right)$

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

Determine the value of bound current and draw the magnetic field of the cylinder for L>>a, for L<<a and for L≈a.

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 $L\approx a$.

Figure 3

Determine the bound current is given by:

Substitute $\hat{\varphi }$for $\overrightarrow{s}$into equation (1).

${\overrightarrow{k}}_b=M\hat{\varphi }$

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.