Skin effect in a metallic conductor plunged in a solenoid
Take 20 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and start searching for the solution.
A detailed solution is then proposed to you.
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A cylindrical solenoid of axis \((Oz)\) and radius \(R_0\) is constituted of \(n\) whorls by meter.
It is travelled by a variable current of intensity \(i(t)=Icos(\omega t)\).
We admit the own magnetic field created by the solenoid is uniform inside (\(r<R_0\)) :
\(\vec B = \mu _0 ni(t)\ \vec u_z\)
And equal to zero outside (\(r>R_0\)).
We also admit the electric field is ortho-radial :
\(\vec E = E\left( {r,t} \right)\ \vec u_\theta\)
Question
Determine the electric field inside the solenoid
Indice
Which equation of Maxwell explains why an electric field appears when the magnetic field depends on time ?
Think about Stoke's theorem.
Solution
Let's use the Faraday's law of induction and Stoke's theorem :
\(\oint_{(C)} {\vec E.d\vec \ell } = - \frac{d}{{dt}}\left( {\iint_{(S)} \vec B .\vec n dS} \right) = - \frac{{d\Phi }}{{dt}}\)
With the hypothesis of the exercise :
\(2\pi rE(r) = - \mu _0 n\frac{{di}}{{dt}}\pi r^2\)
So :
\(\vec E = - \frac{{\mu _0 nr}}{2}\frac{{di}}{{dt}}\vec u_\theta\)
Question
A long and massive cylinder with the same axis \((Oz)\) as the solenoid, of conductivity \(\gamma\), height \(h\) and radius \(R_1<R_0\) is put inside the solenoid.
Determine the density of current \(\vec j\) created by the electric field \(\vec E\).
What is the observable effect associated to these currents ?
Solution
Ohm's local law gives :
\(\vec j = \gamma \vec E = - \frac{{\gamma \mu _0 nr}}{2}\frac{{di}}{{dt}}\vec u_\theta = \frac{{\gamma \mu _0 nr}}{2}I\omega \sin (\omega t)\vec u_\theta\)
Heating due to Joule effect : let's compute the density of power
\(p_J = \vec j.\vec E = \frac{{\gamma \mu _0^2 n^2 r^2 }}{4}\left( {\frac{{di}}{{dt}}} \right)^2\)
Question
Deduce the magnetic field \(\vec B_i\) created by the currents on the axis.
Give the conduction under which this “induced” field is negligible in front of the one created by the solenoid.
Solution
The magnetic field \(\vec B_i\) created by the currents on the axis is along \(\vec u_z\).
We can use Ampère's circuital law to calculate this field.
The field outside the cylinder of radius \(R_1\) is equal to zero.
Using the closed circuit given on the figure, we can write :
\(LB_i=\mu_0 \int_0^{R_1}j(r)Ldr\)
Thus :
\(\vec B_i = \frac{{\gamma \mu _0^2 n}}{2}I\omega \sin (\omega t)\vec u_z \int_0^{R_1 } {rdr}\)
Finally :
\(\vec B_i = \frac{{\gamma \mu _0^2 n}}{4}I\omega R_1^2 \sin (\omega t)\;\vec u_z\)
The ratio of these two magnetic fields is :
\(\frac{{B_i }}{{\mu _0 nI}} \approx \frac{{\gamma \mu _0^{} }}{4}\omega r_1^2 < < 1\;\;\;\;\;\;if\;\;\;\;\;r_1 < < \sqrt {\frac{4}{{\mu _0 \gamma \omega }}} = \sqrt 2 \delta\)
In other words, \(r_1<<\delta\), which is the skin depth.
Question
If this condition is not verified, indicate without any justification the repartition of currents in the cylinder.
Solution
There is a skin effect. The current only exists on the periphery of the cylinder, on a few \(\delta\).