Question

A familiar demonstration of superconductivity (Prob. 7.44) is the levitation of a magnet over a piece of superconducting material. This phenomenon can be analyzed using the method of images. Treat the magnet as a perfect dipole , m a height z above the origin (and constrained to point in the z direction), and pretend that the superconductor occupies the entire half-space below the xy plane. Because of the Meissner effect, B = 0 for $\mathit{Z}\mathbf{\le }\mathbf{0}$, and since B is divergenceless, the normal ( z) component is continuous, so ${\mathbf{B}}_{\mathbf{z}}\mathbf{=}\mathbf{0}$just above the surface. This boundary condition is met by the image configuration in which an identical dipole is placed at - z , as a stand-in for the superconductor; the two arrangements therefore produce the same magnetic field in the region z>0.

(a) Which way should the image dipole point (+ z or -z)?

(b) Find the force on the magnet due to the induced currents in the superconductor (which is to say, the force due to the image dipole). Set it equal to Mg (where M is the mass of the magnet) to determine the height h at which the magnet will "float." [Hint: Refer to Prob. 6.3.]

(c) The induced current on the surface of the superconductor ( xy the plane) can be determined from the boundary condition on the tangential component of B (Eq. 5.76): $\mathbf{B}\mathbf{=}{\mathbf{\mu }}_{\mathbf{0}}\mathbf{(}\mathbf{K}\mathbf{\times }\hat{\mathbf{z}}\mathbf{)}$. Using the field you get from the image configuration, show that

$\mathbf{K}\mathbf{=}\mathbf{-}\frac{\mathbf{3}\mathbf{mrh}}{\mathbf{2}\mathbf{\pi }{\mathbf{(}{\mathbf{r}}^{\mathbf{2}}\mathbf{+}{\mathbf{h}}^{\mathbf{2}}\mathbf{)}}^{\raisebox{1ex}{$\mathbf{5}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{2}$}\right.}}\hat{\mathbf{\varphi }}$

where r is the distance from the origin.

Short answer

(a)The image dipole points towards the downward -z plane.

(b) The height at which the magnet will float is$\frac{1}{2}{\left(\frac{3{\mu }_0{m}^2}{2\pi Mg}\right)}^{\frac{1}{4}}$ .

(c) The value K of is $-\frac{3mh}{2\pi }\frac{r}{{(r^2+h^2)}^{\frac{5}{2}}}\hat{\varphi }$.

Step-by-step solution

Given information

The magnitude of the dipole moment is, m.

The height of the dipole above the origin is,h .

The mass of the magnet is,M .

The equivalent uniform surface current is,K .

Define Meissner effect

The value of the magnetic field inside the superconducting material does not reduce but it cancels out entirely because of the perfect diamagnetism present inside the material.

So when a superconducting material is kept in a magnetic field then the value of the magnetic field inside that particular material would be zero. This effect is described as the “Meissner effect”.

Step 3(a): Determine the direction of the image dipole

When a magnet is levitated over a piece of superconducting material, then this phenomenon can be analyzed using an identical dipole placed at- z plane. The images showing the magnetic field in the region are given below,

From the above figures it is clear that to make the magnetic field parallel to the plane, two monopoles of the same sign are required, so the image dipole points down (-z).

Hence, the image dipole points towards the downward - z plane.

Step 4(b): Determine height at which the magnet will float

The diagram of the two dipoles of magnitude m and -m located at the planes z and -z is given by,

Here, $r_1$and $r_2$are the radial distances of two dipoles at an angle of$\theta$ from the vertical.

Referring to the prob 6.3, the formula for the force between two magnetic dipoles due to the induced currents in the superconductor is given by,

$F=\frac{3{\mu }_0}{2\pi }\frac{m^2}{{\left(2z\right)}^4}$

Equating, the force F = Mg and the height at which the magnet will float is z = h in the expression,

$$\begin{gathered} \mathrm{Mg}=\frac{3{\mathrm{\mu }}_0}{2\mathrm{\pi }}\frac{{\mathrm{m}}^2}{{\left(2\mathrm{h}\right)}^4} \\ {\left(2\mathrm{h}\right)}^4=\frac{3{\mathrm{\mu }}_0{\mathrm{m}}^2}{2\mathrm{\pi Mg}} \\ \mathrm{h}=\frac{1}{2}{\left(\frac{3{\mathrm{\mu }}_0{\mathrm{m}}^2}{2\mathrm{\pi Mg}}\right)}^{\frac{1}{4}} \end{gathered}$$

Hence, the height at which the magnet will float is $\frac{1}{2}{\left(\frac{3{\mu }_0{m}^2}{2\pi Mg}\right)}^{\frac{1}{4}}$.

Step 5(c): Determine the value of K

Similarly, referring to the prob 6.3, the formula for the magnetic field due to the magnetic dipole is given by,

$B_{dip}\left(r\right)=\frac{{\mu }_0}{4\pi }\frac{1}{r^3}\left[3\right(m\cdot \hat{r})\hat{r}-m]$

For two dipoles of magnitude m and -m , the magnetic field is,

$$\begin{gathered} B=\frac{{\mu }_0}{4\pi }\frac{1}{(r_1{)}^3}[3(m\hat{z}\cdot {\hat{r}}_1){\hat{r}}_1-m\hat{z}+3(-m\hat{z}\cdot {\hat{r}}_2){\hat{r}}_2+m\hat{z}] \\ B=\frac{3{\mu }_{0}m}{4\pi (r_1{)}^3}\left[\right(\hat{z}\cdot {\hat{r}}_1){\hat{r}}_1-(\hat{z}\cdot {\hat{r}}_2\left)\hat{r}{}_2\right] \end{gathered}$$

From the diagram, $\hat{z}\cdot {\hat{r}}_1=-\hat{z}\cdot {\hat{r}}_2=cos\theta$,

$$\begin{gathered} \mathrm{B}=\frac{3{\mathrm{\mu }}_0\mathrm{m}}{4\mathrm{\pi }({\mathrm{r}}_1{)}^3}[\mathrm{cos\theta }{\hat{\mathrm{r}}}_1+\mathrm{cos\theta }{\hat{\mathrm{r}}}_2] \\ \mathrm{B}=-\frac{3{\mathrm{\mu }}_0\mathrm{m}}{4\mathrm{\pi }({\mathrm{r}}_1{)}^3}\cos \mathrm{\theta }({\hat{\mathrm{r}}}_1+{\hat{\mathrm{r}}}_2) \end{gathered}$$

Also putting, ${\hat{r}}_1+{\hat{r}}_2=2sin\theta \hat{r}$in the expression,

$$\begin{gathered} B=-\frac{3{\mu }_{0}m}{4\pi (r_1{)}^3}cos\theta (2sin\theta \hat{r}) \\ B=-\frac{3{\mu }_{0}m}{2\pi (r_1{)}^3}sin\theta cos\theta \hat{r} \end{gathered}$$

Using right angled triangle formula,$\mathrm{sin\theta }=\frac{\mathrm{r}}{{\mathrm{r}}_1},\mathrm{cos\theta }=\frac{\mathrm{h}}{{\mathrm{r}}_1},{\mathrm{r}}_1=\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}$,

Putting the values and solving,

$$\begin{gathered} B=-\frac{3{\mu }_{0}mrh}{2\pi (r_1{)}^5}\times \hat{r} \\ B=-\frac{3{\mu }_{0}mrh}{2\pi {\left(\sqrt{r^2+h^2}\right)}^5}\hat{r} \\ B=-\frac{3{\mu }_{0}mrh}{2\pi {(r^2+h^2)}^{{\displaystyle \frac{5}{2}}}}\hat{r} \end{gathered}$$

It is given that, $B={\mu }_0(K\times \hat{z})$,

Cross multiplying both sides by $\hat{z}$,

$$\begin{gathered} \hat{z}\times B={\mu }_0\hat{z}\times \left(K\times \hat{z}\right) \\ \hat{z}\times B={\mu }_0\left[K-\hat{z}\times \left(K\times \hat{z}\right)\right] \end{gathered}$$

Here, $K\times \hat{z}=0$because the surface current is in the xy plane.

$$\begin{gathered} \hat{z}\times B={\mu }_{0}K-0 \\ \hat{z}\times B={\mu }_{0}K \\ K=\frac{1}{{\mu }_0}\left(\hat{z}\times B\right) \end{gathered}$$

Putting the value of ,

$$\begin{gathered} K=\frac{1}{{\mu }_0}\left\{\hat{z}\times \left(-\frac{3{\mu }_{0}mrh}{2\pi {(r^2+h^2)}^{\frac{5}{2}}}\hat{r}\right)\right\} \\ K=-\frac{3mh}{2\pi }\frac{r}{{(r^2+h^2)}^{\frac{5}{2}}}\left(\hat{z}\times \hat{r}\right) \\ K=-\frac{3mh}{2\pi }\frac{r}{{(r^2+h^2)}^{\frac{5}{2}}}\hat{\varphi } \end{gathered}$$

Hence, the value of K is $-\frac{3mh}{2\pi }\frac{r}{{(r^2+h^2)}^{\frac{5}{2}}}\hat{\varphi }$.