Q7.46P - Introduction to Electrodynamics

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?

Step-by-Step Solution

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 1: 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.

Step 2: 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)

Step 3: Calculate the height above the surface.

$m_1$

Let’s assume moment of dipole is $m_1$ 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 $Mg$.