Question

You are asked to referee a grant application, which proposes to determine whether the magnetization of iron is due to "Ampere" dipoles (current loops) or "Gilbert" dipoles (separated magnetic monopoles). The experiment will involve a cylinder of iron (radius $\mathbf{R}$and length $\mathbf{L}\mathbf{=}\mathbf{10}\mathbf{R}$), uniformly magnetized along the direction of its axis. If the dipoles are Ampere-type, the magnetization is equivalent to a surface bound current ${\mathbf{K}}_{\mathbf{b}}\mathbf{=}\mathbf{M}\widehat{\mathbf{\varphi }}$if they are Gilbert-type, the magnetization is equivalent to surface monopole densities ${\mathbf{\sigma }}_{\mathbf{b}}\mathbf{=}\mathbf{\pm }\mathbf{M}$at the two ends. Unfortunately, these two configurations produce identical magnetic fields, at exterior points. However, the interior fields are radically different-in the first case $\mathbf{B}$is in the same general direction as $\mathbf{M}$, whereas in the second it is roughly opposite to $\mathbf{M}$. The applicant proposes to measure this internal field by carving out a small cavity and finding the torque on a tiny compass needle placed inside.

Assuming that the obvious technical difficulties can be overcome, and that the question itself is worthy of study, would you advise funding this experiment? If so, what shape cavity would you recommend? If not, what is wrong with the proposal?

Short answer

The fields are equal in both the proposed model for spherical, needle and water shaped cavity. Therefore, I advise to found it.

Step-by-step solution

Write the given data from the question.

The radius of the cylinder is $R$.

The length of the cylinder,$L=10R$

The magnetic field is $B$.

The surface bound current is $K_b=M\widehat{\varphi }$

The magnetization is $M$.

Determine the formulas to calculate the field expression.

For Ampere’s dipoles

The expression for the field inside the material is given as follows.

${\mathbf{B}}_{\mathbf{0}}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{M}$

For Gilbert’s dipoles

The field inside the material is equivalent to the magnetic field at the midpoint between the two-point charges.

${\mathbf{B}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$

The expression of field of magnetized sphere is given as follows.

$\mathit{B}\mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}{\mathit{\mu }}_{\mathbf{0}}\mathit{M}$

Calculate the expression for the field.

Calculate the expression for the filed, if the cavity is spherical in the shape.

$\mathrm{B}={\mathrm{B}}_0-\frac{2}{3}{\mathrm{\mu }}_0\mathrm{M}$

Substitute${\mathrm{\mu }}_0\mathrm{M}$ for${\mathrm{B}}_0$into above equation.

$\begin{array}{l}\mathrm{B}={\mathrm{\mu }}_0\mathrm{M}-\frac{2}{3}{\mathrm{\mu }}_0\mathrm{M}\\ \mathrm{B}=\frac{1}{3}{\mathrm{\mu }}_0\mathrm{M}\end{array}$

Therefore, the field inside the spherical cavity is $\frac{1}{3}{\mathrm{\mu }}_0\mathrm{M}$.

Calculate the expression for the filed, if the cavity is needle type in the shape.

$\mathrm{B}={\mathrm{B}}_0-{\mathrm{\mu }}_0\mathrm{M}$

Substitute${\mu }_{0}M$for$B_0$into above equation.

$\begin{array}{l}B={\mu }_{0}M-{\mu }_{0}M\\ B=0\end{array}$

Therefore, the field inside the needle type cavity is $0$.

Calculate the expression for the filed, if the cavity is water shaped.

$B={\mu }_{0}M$

Therefore, the field inside the water shaped cavity is ${\mathrm{\mu }}_0\mathrm{M}$.

Consider the dipole to be Gilberts dipoles.

The field inside the material is equivalent to the magnetic field at the midpoint between the two-point charges.

$B_0=0$

The field at the centre of the cavity, if cavity is spherical in shape.

$\mathrm{B}={\mathrm{B}}_0+\frac{1}{3}{\mathrm{\mu }}_0\mathrm{M}$

Substitute$0$for$B_0$into above equation.

$\begin{array}{l}B=0+\frac{1}{3}{\mu }_{0}M\\ B=\frac{1}{3}{\mu }_{0}M\end{array}$

Therefore, the field inside the spherical cavity is $\frac{1}{3}{\mu }_{0}M$.

The field at the centre of the cavity, if cavity is needle type shape.

$B=B_0$

Substitute$0$for$B_0$into above equation.

$B=0$

Therefore, the field inside the needle type cavity is$0$.

The field at the centre of the cavity, if cavity is water shaped.

$B=B_0+{\mu }_{0}M$

Substitute$0$for$B_0$into above equation.

$\begin{array}{l}B=0+{\mu }_{0}M\\ B={\mu }_{0}M\end{array}$

Therefore, the field inside the water shaped cavity is ${\mu }_{0}M$.

Since the fields are equal in both the proposed model for spherical, needle and water shaped cavity. Therefore, I advise to found it.