# In a perfect conductor, the conductivity is infinite, so E = 0, and any net charge resides on the...

## Question:

In a perfect conductor, the conductivity is infinite, so E = 0, and any net charge resides on the surface. a) Show that the magnetic field is constant (dB/dt=0) 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 is in fact zero. c) Show that the current in a superconductor is confined to the surface. d) Superconductivity is lost above a certain critical temperature (Tc), which varies from one material to another. Suppose you had a sphere (radius b) above its critical temperature, and you held it in a uniform magnetic field B0ez while cooling it below Tc. Find the induced surface current density K, as a function of the polar angle theta.

## Faraday's Law:

Faraday's law states that the magnitude of the electric field is given as the rate of change of magnetic flux and the expression for the Faraday's law can be given by,

{eq}\triangledown \times E=-\frac{\delta B}{\delta t} {/eq}

- E is the electric field.
- B is the magnetic field.

## Answer and Explanation:

(a) Using Faraday's laws,

{eq}\triangledown \times E=-\frac{\delta B}{\delta t} {/eq}

In a perfect conductor, E is zero.

Substituting 0 for E,

{eq}\frac{\delta B}{\delta t}=0 {/eq}

That means the magnetic field(b) is constant.

(b) Using the integral form of Faraday's law,

{eq}\oint E.dl=-\frac{d\phi }{dt} {/eq}

Here {eq}\phi {/eq} is the representation of magnetic flux.

In a perfect conductor, E=0

Substitute 0 for E,

{eq}\frac{d\phi }{dt}=0 {/eq}

thus, {eq}\phi {/eq} is constant.

Thus the magnetic flux is constant in the loop.

(c) In a superconductor, the magnetic field is zero,

Using the following equation of Ampere-Maxwell law,

{eq}\triangledown \times B=\mu _{0}J+\mu _{0}\epsilon _{0}\frac{\delta E}{\delta t} {/eq}

In the given equation,

J is the current density, {eq}\epsilon _{0} {/eq} is permittivity, {eq}\mu _{0} {/eq} is permeability,

If{eq}E=0 {/eq}, then {eq}B=0 {/eq} thus {eq}J=0 {/eq}

Thus, the current must be confined to the surface.

(d) Given the radius of the sphere is b.

Inside the shell, the magnetic field will be uniform and it can be expressed as,

{eq}B=\frac{2}{3}\mu _{0}\sigma \omega b\hat{z} {/eq}

But to cancel the magnetic field inside the value of {eq}B_{0} {/eq} is,

{eq}B_{0}=-\frac{2}{3}\mu _{0}\sigma \omega b\hat{z} {/eq}

{eq}\sigma \omega b=-\frac{3}{2}\frac{B_{0}}{\mu _{0}} {/eq}

Surface density K can be given as,

{eq}K=\sigma v {/eq}

{eq}K=\sigma v=\sigma \omega bsin\theta \hat{\phi } {/eq}

Substituting the value, {eq}-\frac{3}{2}\frac{B_{0}}{\mu _{0}} {/eq} for {eq}\sigma \omega b {/eq}

{eq}K=-\frac{3}{2}\frac{B_{0}}{\mu _{0}}sin\theta \hat{\phi } {/eq}