Uniform rotation of a cylinder evenly charged in volume

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A detailed solution is then proposed to you.

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Let \((C)\) be a cylinder of revolution around the axis \((Oz)\), of radius \(a\).

Its length is very big compared to \(a\).

\((C)\) is evenly charged with a volume density \(\rho\).

It is set in motion to rotate around \((Oz)\) axis, with an angular speed \(\omega\).

This rotation does not depend on time, and it does not affect the repartition of charges in \((C)\).

Question

Find the electric field \(\vec E\).

Indice

Use Gauss'law.

Solution

Let's use Gauss' law : (the electric field is radial)

We choose a cylinder of radius \(r\) and axis (Oz).

  • For \(r>a\) :

    \(2\pi rhE(r) = \frac{1}{{{\varepsilon _0}}}\pi {a^2}h\rho \;\;\;\;\;so\;\;\;\;\;E(r) = \frac{\rho }{{2{\varepsilon _0}}}\frac{{{a^2}}}{r}\)

  • For \(r<a\) :

    \(2\pi rhE(r) = \frac{1}{{{\varepsilon _0}}}\pi {r^2}h\rho \;\;\;\;\;so\;\;\;\;\;E(r) = \frac{\rho }{{2{\varepsilon _0}}}r\)

We have verified that the electric field in continuous when crossing the cylinder (In \(r=a\))

Question

Find the magnetic field \(\vec B\) at any point in space.

Indice

The currents are ortho-radial. Use the same reasoning as for a classical solenoid and use Ampere's circuital law.

Solution

Let's use Ampere's circuital law :

The magnetic field is oriented as the axis of the solenoid, and we know it is equal to zero outside.

Let us choose a rectangular loop of lenght \(L\).

One of its sides is parallel to the axis and is inside the solenoid, whereas the other one is outside the solenoid.

At a distance \(r'\) from the (Oz) axis, the current density is given by :

\(\vec j = \rho \vec v = \rho r' \omega \;\vec u_{\theta}\)

Thus, for \(r<a\), Ampere's circuital law gives :

\(LB(r)=\mu_0 \iint \vec j . \vec u_{\theta} dS =\mu_0 \int_r^{a} \vec j . \vec u_{\theta} Ldr' \)

Finally :

\(B(r) = {\mu _0}\int_r^a {\rho \omega r'dr'} = \frac{{{\mu _0}\rho \omega }}{2}({a^2} - {r^2})\)