A thin conducting disk of thickness h, radius a and conductivity \sigma is placed in a uniform...

Question:

A thin conducting disk of thickness h, radius a and conductivity {eq}\sigma {/eq} is placed in a uniform alternating magnetic field {eq}B =B_0 sin(\omega t) {/eq} parallel to the axis of the disk.

Find the induced current density as a function of distance from the axis of the disk. What is the direction of this current?

Induced current:

The induced current is generated from the changing magnetic field. Whenever there is a change in magnetic flux over a wire, the current gets induced in it. The quantity of current that passes through a certain area is termed as current density.

Answer and Explanation:

Given data

  • The thickness of disk is h
  • The radius of disk is a
  • The conductivity of disk is {eq}\sigma {/eq} .
  • The alternating magnetic field is {eq}B = {B_o}\sin \left( {\omega t} \right) {/eq}.


The expression for the current density is given as,

{eq}J = \dfrac{I}{A}......\left( 1 \right) {/eq}

The expression for the induced emf is given as,

{eq}\varepsilon = \left| { - \dfrac{{d\phi }}{{dt}}} \right|......\left( 2 \right) {/eq}


The expression for the flux is given as,

{eq}\phi = BA......\left( 3 \right) {/eq}

Consider a circular ring of radius r and width dr.

Substituting the values in equation 3,

{eq}\begin{align*} \phi &= B.ds\\ \phi &= 2\pi rdr \times {B_o}\sin \omega t \end{align*} {/eq}


Substituting the values in equation 2,

{eq}\begin{align*} \varepsilon &= 2\pi rdr \times \dfrac{d}{{dt}}{B_o}\sin \omega t\\ \varepsilon &= 2\pi rdr \times {B_o}\omega \cos \omega t \end{align*} {/eq}


As the resistance is given as,

{eq}\begin{align*} R &= \dfrac{{\rho l}}{A}\\ R &= \dfrac{l}{{\sigma A}}\\ R &= \dfrac{{2\pi r}}{{\sigma hdr}} \end{align*} {/eq}


The induced current is given as,

{eq}\begin{align*} I &= \dfrac{\varepsilon }{R}\\ I &= \dfrac{{2\pi rdr \times {B_o}\omega \cos \omega t}}{{\dfrac{{2\pi r}}{{\sigma hdr}}}}\\ I &= {B_o}\omega h\sigma {\left( {dr} \right)^2}\cos \omega t \end{align*} {/eq}


Substituting the values in equation 1,

{eq}\begin{align*} J &= \dfrac{{{B_o}\omega h\sigma {{\left( {dr} \right)}^2}\cos \omega t}}{{hdr}}\\ J &= {B_o}\omega \sigma \left( {dr} \right)\cos \omega t \end{align*} {/eq}


Thus, the induced current density is {eq}{B_o}\omega \sigma \left( {dr} \right)\cos \omega t {/eq}.


The induced current will be in the clockwise direction as it will resist the variation in the magnetic field.


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