91Ó°ÊÓ

The magnetic dipole is levitating above the infinite superconducting plane.

The image of the dipole at distance$h$ from the negative z axis.

The expression for magnetic field of image dipole moment is given as follows.

$B\left(z\right)=\frac{{\mathrm{μ}}_0}{4\mathrm{π}}\frac{1}{{(h+z)}^3}[3(m×\widehat{z})\widehat{z}-m]$ …… (1)

Here $m_2$is the magnetic dipole moment.

The expression to calculate the torque on the dipole moment is given as follows.

$N=m×B$ …… (2)

The expression to calculate the force on the magnetic dipole moment is given as follows.

$F=∇(m_1×B)$ …â¶Ä¦.(3)

$m_1$

Let’s assume moment of dipole is localid="1658401259550" $m_1$and angle made by the dipole with z axis is $\mathrm{θ}$.

The magnetic dipole moment is given by,

$m_1=m\sin \mathrm{θ}\widehat{x}+m\cos \mathrm{θ}\widehat{z}$

The magnetic dipole moment$m_2$is given by,

$m_2=m\sin \mathrm{θ}\widehat{x}−m\cos \mathrm{θ}\widehat{z}$

The magnetic field of image dipole momentis given by,

$B\left(z\right)=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{{(h+z)}^3}[3(m_2â‹\dots \widehat{z})\widehat{z}−m_2]$

The force on the magnetic dipole moment $m_1$is given by,

$N=m_1×B\left(z\right)$

Substitute $\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{{(h+z)}^3}[3(m_2â‹\dots \widehat{z})\widehat{z}−m_2]$for$B\left(z\right)$into above equation.

$N=m_1×\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{{(h+z)}^3}[3(m_2â‹\dots \widehat{z})\widehat{z}−m_2]$

Substitute$h$for $z$into above equation.

$N=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{{\left(2h\right)}^3}[3(m_2â‹\dots \widehat{z})(m_1×\widehat{z})−m_2×m_1]$

Substitute$m\sin \mathrm{θ}\widehat{x}+m\cos \mathrm{θ}\widehat{z}$ for $m_1$and $m\sin \mathrm{θ}\widehat{x}−m\cos \mathrm{θ}\widehat{z}$for $m_2$into above equation.

$\begin{array}{l}N=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{{\left(2h\right)}^3}\left\{\begin{array}{l}3[(m\sin \mathrm{θ}\widehat{x}−m\cos \mathrm{θ}\widehat{z})â‹\dots \widehat{z}][(m\sin \mathrm{θ}\widehat{x}+m\cos \mathrm{θ}\widehat{z})×\widehat{z}]\\ −(m\sin \mathrm{θ}\widehat{x}−m\cos \mathrm{θ}\widehat{z})×(m\sin \mathrm{θ}\widehat{x}+m\cos \mathrm{θ}\widehat{z})\end{array}\right\}\\ N=\frac{{\mathrm{μ}}_0}{4\mathrm{Ï€}}\frac{1}{8h^3}[3(−m\cos \mathrm{θ})(−m\sin \mathrm{θ})\widehat{y}−2m^2\cos \mathrm{θ}\sin \mathrm{θ}\widehat{y}]\\ N=\frac{{\mathrm{μ}}_0}{32h^3\mathrm{Ï€}}[3m^2\cos \mathrm{θ}\sin \mathrm{θ}\widehat{y}−2m^2\cos \mathrm{θ}\sin \mathrm{θ}\widehat{y}]\\ N=\frac{{\mathrm{μ}}_0}{32h^3\mathrm{Ï€}}{m}^2\cos \mathrm{θ}\sin \mathrm{θ}\widehat{y}\end{array}$

The torque would be zero at$\mathrm{θ}=0,\text{ }\mathrm{π},\mathrm{π}/2$ . But$\mathrm{θ}=0$and$\mathrm{π}/\text{2}$is unstable.

Therefore,$\mathrm{θ}=\frac{\mathrm{π}}{2}$ which is parallel to the surface.

The force on the magnetic dipole is given by,

$F=∇(m_1â‹\dots B)$

Substitute$\frac{−{\mathrm{μ}}_{0}m}{4\mathrm{π}{(h+z)}^3}$ for $B$and $m\sin \mathrm{θ}\widehat{x}+m\cos \mathrm{θ}\widehat{z}$for$m_1$into above equation.

$\begin{array}{l}F=∇((m\sin \mathrm{θ}\widehat{x}+m\cos \mathrm{θ}\widehat{z})â‹\dots \frac{−{\mathrm{μ}}_{0}m}{4\mathrm{Ï€}{(h+z)}^3})\\ F={\frac{3{\mathrm{μ}}_0{m}^3}{4\mathrm{Ï€}{(h+z)}^4}|}_{z=h}\widehat{z}\\ F=\frac{3{\mathrm{μ}}_0{m}^3}{4\mathrm{Ï€}{\left(2h\right)}^4}\widehat{z}\end{array}$

At the equilibrium, the force is balanced by the weight that is $Mg$.