91影视

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.

$B\left(z\right)=\frac{{渭}_0}{4蟺}\frac{1}{{(h+z)}^3}[3(m脳\widehat{z})\widehat{z}-m]$ 鈥︹�� (1)

Here $m_2$is the magnetic dipole moment.

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

$N=m脳B$ 鈥︹�� (2)

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

$F=鈭�(m_1脳B)$ 鈥︹赌�.(3)

$m_1$

Let鈥檚 assume moment of dipole is localid="1658401259550" $m_1$and angle made by the dipole with z axis is $胃$.

The magnetic dipole moment is given by,

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

The magnetic dipole moment$m_2$is given by,

$m_2=m\sin 胃\widehat{x}鈭�m\cos 胃\widehat{z}$

The magnetic field of image dipole momentis given by,

$B\left(z\right)=\frac{{渭}_0}{4蟺}\frac{1}{{(h+z)}^3}[3(m_2鈰�\widehat{z})\widehat{z}鈭�m_2]$

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

$N=m_1脳B\left(z\right)$

Substitute $\frac{{渭}_0}{4蟺}\frac{1}{{(h+z)}^3}[3(m_2鈰�\widehat{z})\widehat{z}鈭�m_2]$for$B\left(z\right)$into above equation.

$N=m_1脳\frac{{渭}_0}{4蟺}\frac{1}{{(h+z)}^3}[3(m_2鈰�\widehat{z})\widehat{z}鈭�m_2]$

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

$N=\frac{{渭}_0}{4蟺}\frac{1}{{\left(2h\right)}^3}[3(m_2鈰�\widehat{z})(m_1脳\widehat{z})鈭�m_2脳m_1]$

Substitute$m\sin 胃\widehat{x}+m\cos 胃\widehat{z}$ for $m_1$and $m\sin 胃\widehat{x}鈭�m\cos 胃\widehat{z}$for $m_2$into above equation.

$\begin{array}{l}N=\frac{{渭}_0}{4蟺}\frac{1}{{\left(2h\right)}^3}\left\{\begin{array}{l}3[(m\sin 胃\widehat{x}鈭�m\cos 胃\widehat{z})鈰�\widehat{z}][(m\sin 胃\widehat{x}+m\cos 胃\widehat{z})脳\widehat{z}]\\ 鈭�(m\sin 胃\widehat{x}鈭�m\cos 胃\widehat{z})脳(m\sin 胃\widehat{x}+m\cos 胃\widehat{z})\end{array}\right\}\\ N=\frac{{渭}_0}{4蟺}\frac{1}{8h^3}[3(鈭�m\cos 胃)(鈭�m\sin 胃)\widehat{y}鈭�2m^2\cos 胃\sin 胃\widehat{y}]\\ N=\frac{{渭}_0}{32h^3蟺}[3m^2\cos 胃\sin 胃\widehat{y}鈭�2m^2\cos 胃\sin 胃\widehat{y}]\\ N=\frac{{渭}_0}{32h^3蟺}{m}^2\cos 胃\sin 胃\widehat{y}\end{array}$

The torque would be zero at$胃=0,\text{鈥�}蟺,蟺/2$ . But$胃=0$and$蟺/\text{2}$is unstable.

Therefore,$胃=\frac{蟺}{2}$ which is parallel to the surface.

The force on the magnetic dipole is given by,

$F=鈭�(m_1鈰�B)$

Substitute$\frac{鈭�{渭}_{0}m}{4蟺{(h+z)}^3}$ for $B$and $m\sin 胃\widehat{x}+m\cos 胃\widehat{z}$for$m_1$into above equation.

$\begin{array}{l}F=鈭�((m\sin 胃\widehat{x}+m\cos 胃\widehat{z})鈰�\frac{鈭�{渭}_{0}m}{4蟺{(h+z)}^3})\\ F={\frac{3{渭}_0{m}^3}{4蟺{(h+z)}^4}|}_{z=h}\widehat{z}\\ F=\frac{3{渭}_0{m}^3}{4蟺{\left(2h\right)}^4}\widehat{z}\end{array}$

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