Q44P - Introduction to Electrodynamics

Question

In a perfect conductor, the conductivity is infinite, so $\mathbf{E}\mathbf{=}\mathbf{0}$ (Eq. 7.3), and any net charge resides on the surface (just as it does for an imperfect conductor, in electrostatics).

(a) Show that the magnetic field is constant $\left(\frac{\partial B}{\partial t}=0\right)$, inside a perfect conductor.

(b) Show that the magnetic flux through a perfectly conducting loop is constant.

A superconductor is a perfect conductor with the additional property that the (constant) B inside is in fact zero. (This "flux exclusion" is known as the Meissner effect.)

(c) Show that the current in a superconductor is confined to the surface.

(d) Superconductivity is lost above a certain critical temperature $\mathbf{\left(}{\mathbf{T}}_{\mathbf{c}}\mathbf{\right)}$, which varies from one material to another. Suppose you had a sphere (radius ) above its critical temperature, and you held it in a uniform magnetic field ${\mathbf{B}}_{\mathbf{0}}\hat{\mathbf{z}}$ while cooling it below ${\mathbf{T}}_{\mathbf{c}}$ . Find the induced surface current density K, as a function of the polar angle $\mathit{\theta }$.

Step-by-Step Solution

Answer

(a) The magnetic field inside the conductor is 0.

(b) The magnetic field inside the conducting loop is constant.

(c) It is proved that the current in the superconductor is confined to the surface.

(d) The induced surface current density is $k=-\frac{3B_0}{2\mu }\sin \theta \widehat{\varphi .}$

Step 1: Faraday’s law

Based on this law whenever a conductor is kept inside a varying magnetic field then it experiences a force known as ‘electro motive force (emf)’ as well as a certain current is induced.

 

The value of emf generated on a conducting coil relies upon the change of magnetic flux as well as the number of turns of the coil.

Step 2(a): Magnetic field inside a perfect conductor.

Applying Faraday’s law, the expression for the magnetic field inside a perfect conductor is given by,

 

$\nabla \times E=-\frac{\partial B}{\partial t}$

 

Here, E is the electric field and B is the magnetic field inside a perfect conductor.

Putting E=0 in the expression,

 

$$\begin{gathered} \nabla \times E=-\frac{\partial B}{\partial t} \\ \nabla \times 0=-\frac{\partial B}{\partial t} \\ \frac{\partial B}{\partial t}=0 \end{gathered}$$

 

Hence, the magnetic field is constant inside a perfect conductor.

Step 3(b): Magnetic flux through a perfectly conducting loop

Using Faraday’s law, the integral formula for the magnetic flux through a perfectly conducting loop is given by,

 

$\int E.dl=-\frac{d\Phi }{dt}$

Here, E is the electric field and $\Phi$ is the magnetic flux through a perfectly conducting loop.

Putting $E=0$ in the expression,

 

$$\begin{gathered} \int 0.dl=-\frac{d\Phi }{dt} \\ -\frac{d\Phi }{dt}=0 \\ \frac{d\Phi }{dt}=0 \end{gathered}$$

 

Hence, the magnetic flux through a perfectly conducting loop is constant.

Step 4(c): The current in a superconductor

The generalized form of Ampere-Maxwell formula is given by,

 

$\nabla \times B={\mu }_{0}J+{\mu }_0 {\in }_0\frac{\partial E}{\partial t}$

 

Here, E represents the electric field, ${\mu }_0$ is the permeability of free space, J is the current in the superconductor and $\frac{\partial E}{\partial t}$ is the change in electric field.

Putting $E=0$ and $B=0$ in expression,

 

$$\begin{gathered} \nabla \times 0={\mu }_{0}J+{\mu }_0 {\in }_0 \times 0 \\ {\mu }_{0}J=0 \\ J=0 \end{gathered}$$

 

Hence, the current in a superconductor is confined to the surface.

Step 5(d): The induced surface current density

The expression for the uniform magnetic field generated inside a rotating shell in polar form is given by,

 

$$\begin{gathered} B=\nabla \times A \\ B=\frac{2{\mu }_{0}R\omega \delta }{3}(\cos \theta \hat{r}-\sin \theta \hat{\theta }) \\ B=\frac{2}{3}{\mu }_0\delta R\omega \hat{z} \\ B=\frac{2}{3}{\mu }_0\delta R\omega \end{gathered}$$

 

Putting the value of radius R=a in the expression,

 

$$\begin{gathered} B=\frac{2}{3}{\mu }_0\delta \omega a\hat{Z} \\ \delta \omega a=-\frac{2B_0}{3{\mu }_0} \end{gathered}$$

The formula for the induced surface current density of the sphere is given by,

 

$K=\delta \nu$

 

Here, $\delta$ is the surface charge density and $\nu$ is the velocity of the charge.

Putting the value of charge velocity $\nu =\omega \times a \sin \theta \hat{\varphi }$ in the expression,

 

$$\begin{gathered} K=\delta \omega a\sin \theta \hat{\varphi } \\ K=-\frac{3B_0}{2{\mu }_0}\sin \theta \hat{\varphi } \end{gathered}$$
 

Hence, the induced surface current density is $k=-\frac{3B_0}{2{\mu }_0}\sin \theta \hat{\varphi }$.