Question

If a magnetic dipole levitating above an infinite superconducting plane (Pro b. 7 .45) is free to rotate, what orientation will it adopt, and how high above the surface will it float?

Short answer

The orientation is parallel to the surface and height is above which it will float is $\frac{1}{2}{\left(\frac{3{\mu }_0{m}^3}{4\pi Mg}\right)}^{\frac{1}{4}}$.

Step-by-step solution

Write the given data from the question.

The magnetic dipole is levitating above the infinite superconducting plane.

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

Determine the formula to calculate the height above the surface.

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

$B\left(z\right)=\frac{{\mu }_0}{4\pi }\frac{1}{{(h+z)}^3}[3(m\times \widehat{z})\widehat{z}-m]$ …… (1)

Here $m_2$is the magnetic dipole moment.

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

$N=m\times B$ …… (2)

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

$F=\nabla (m_1\times B)$ …….(3)

Calculate the height above the surface.

$m_1$

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

The magnetic dipole moment is given by,

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

The magnetic dipole moment$m_2$is given by,

$m_2=m\sin \theta \widehat{x}-m\cos \theta \widehat{z}$

The magnetic field of image dipole momentis given by,

$B\left(z\right)=\frac{{\mu }_0}{4\pi }\frac{1}{{(h+z)}^3}[3(m_2\cdot \widehat{z})\widehat{z}-m_2]$

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

$N=m_1\times B\left(z\right)$

Substitute $\frac{{\mu }_0}{4\pi }\frac{1}{{(h+z)}^3}[3(m_2\cdot \widehat{z})\widehat{z}-m_2]$for$B\left(z\right)$into above equation.

$N=m_1\times \frac{{\mu }_0}{4\pi }\frac{1}{{(h+z)}^3}[3(m_2\cdot \widehat{z})\widehat{z}-m_2]$

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

$N=\frac{{\mu }_0}{4\pi }\frac{1}{{\left(2h\right)}^3}[3(m_2\cdot \widehat{z})(m_1\times \widehat{z})-m_2\times m_1]$

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

$\begin{array}{l}N=\frac{{\mu }_0}{4\pi }\frac{1}{{\left(2h\right)}^3}\left\{\begin{array}{l}3[(m\sin \theta \widehat{x}-m\cos \theta \widehat{z})\cdot \widehat{z}][(m\sin \theta \widehat{x}+m\cos \theta \widehat{z})\times \widehat{z}]\\ -(m\sin \theta \widehat{x}-m\cos \theta \widehat{z})\times (m\sin \theta \widehat{x}+m\cos \theta \widehat{z})\end{array}\right\}\\ N=\frac{{\mu }_0}{4\pi }\frac{1}{8h^3}[3(-m\cos \theta )(-m\sin \theta )\widehat{y}-2m^2\cos \theta \sin \theta \widehat{y}]\\ N=\frac{{\mu }_0}{32h^3\pi }[3m^2\cos \theta \sin \theta \widehat{y}-2m^2\cos \theta \sin \theta \widehat{y}]\\ N=\frac{{\mu }_0}{32h^3\pi }{m}^2\cos \theta \sin \theta \widehat{y}\end{array}$

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

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

The force on the magnetic dipole is given by,

$F=\nabla (m_1\cdot B)$

Substitute$\frac{-{\mu }_{0}m}{4\pi {(h+z)}^3}$ for $B$and $m\sin \theta \widehat{x}+m\cos \theta \widehat{z}$for$m_1$into above equation.

$\begin{array}{l}F=\nabla ((m\sin \theta \widehat{x}+m\cos \theta \widehat{z})\cdot \frac{-{\mu }_{0}m}{4\pi {(h+z)}^3})\\ F={\frac{3{\mu }_0{m}^3}{4\pi {(h+z)}^4}|}_{z=h}\widehat{z}\\ F=\frac{3{\mu }_0{m}^3}{4\pi {\left(2h\right)}^4}\widehat{z}\end{array}$

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