91影视

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 .

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 鈥淢eissner effect鈥�.

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.

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$胃$ 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,

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

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

$$\begin{gathered} \mathrm{Mg}=\frac{3{\mathrm{渭}}_0}{2\mathrm{蟺}}\frac{{\mathrm{m}}^2}{{\left(2\mathrm{h}\right)}^4} \\ {\left(2\mathrm{h}\right)}^4=\frac{3{\mathrm{渭}}_0{\mathrm{m}}^2}{2\mathrm{蟺惭驳}} \\ \mathrm{h}=\frac{1}{2}{\left(\frac{3{\mathrm{渭}}_0{\mathrm{m}}^2}{2\mathrm{蟺惭驳}}\right)}^{\frac{1}{4}} \end{gathered}$$

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

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

$B_{dip}\left(r\right)=\frac{{渭}_0}{4蟺}\frac{1}{r^3}\left[3\right(m鈰�\hat{r})\hat{r}-m]$

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

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

From the diagram, $\hat{z}鈰�{\hat{r}}_1=-\hat{z}鈰�{\hat{r}}_2=cos胃$,

$$\begin{gathered} \mathrm{B}=\frac{3{\mathrm{渭}}_0\mathrm{m}}{4\mathrm{蟺}({\mathrm{r}}_1{)}^3}[\mathrm{肠辞蝉胃}{\hat{\mathrm{r}}}_1+\mathrm{肠辞蝉胃}{\hat{\mathrm{r}}}_2] \\ \mathrm{B}=-\frac{3{\mathrm{渭}}_0\mathrm{m}}{4\mathrm{蟺}({\mathrm{r}}_1{)}^3}\cos \mathrm{胃}({\hat{\mathrm{r}}}_1+{\hat{\mathrm{r}}}_2) \end{gathered}$$

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

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

Using right angled triangle formula,$\mathrm{蝉颈苍胃}=\frac{\mathrm{r}}{{\mathrm{r}}_1},\mathrm{肠辞蝉胃}=\frac{\mathrm{h}}{{\mathrm{r}}_1},{\mathrm{r}}_1=\sqrt{{\mathrm{r}}^2+{\mathrm{h}}^2}$,

Putting the values and solving,

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

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

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

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

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

Putting the value of ,

$$\begin{gathered} K=\frac{1}{{渭}_0}\left\{\hat{z}脳\left(-\frac{3{渭}_{0}mrh}{2蟺{(r^2+h^2)}^{\frac{5}{2}}}\hat{r}\right)\right\} \\ K=-\frac{3mh}{2蟺}\frac{r}{{(r^2+h^2)}^{\frac{5}{2}}}\left(\hat{z}脳\hat{r}\right) \\ K=-\frac{3mh}{2蟺}\frac{r}{{(r^2+h^2)}^{\frac{5}{2}}}\widehat{蠒} \end{gathered}$$

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