Electromagnetic field of cylinder

In classical electromagnetism, electromagnetic field of cylinder is considered as the field of one of the simplest geometric bodies. The solutions for the components of the electromagnetic field for the case of a stationary and rotating cylinder are not much more complicated than the corresponding solutions for a ball.

When a uniformly charged cylinder rotates with a constant angular velocity, the field is stationary and does not depend on time. In this case, for electric scalar potential $$  \varphi $$ and for magnetic vector potential $$ \mathbf A $$ from Maxwell's equations follow equations:


 * $$ \Delta \varphi = -\frac {\gamma \rho_{0q}}{\varepsilon_0 }, \qquad (1) $$


 * $$ \Delta \mathbf A = -\frac {\gamma \rho_{0q}}{\varepsilon_0 c^2} \mathbf v, \qquad (2) $$

where $$ \Delta $$ is the Laplace operator; $$ \gamma $$ is the Lorentz factor; $$ \rho_{0q} $$ is the invariant charge density of the cylinder matter; $$ \varepsilon_0 $$ is the electric constant; $$ c $$ is the speed of light; $$ \mathbf v $$ is the linear speed of rotation of an arbitrary point taken in the volume of the cylinder.

Long stationary cylinder
In a fixed cylinder, the Lorentz factor of charged particles of matter is $$ \gamma = 1 $$, if the proper chaotic motion of these particles is not taken into account. It is convenient to express the Laplacian in (1) in cylindrical coordinates $$ \rho \, \phi \, z $$. In a sufficiently long cylinder, one can neglect the edge effects that are significant near the ends of the cylinder, and assume that the field mainly depends only on the coordinate $$ \rho $$. In this approximation, the electric potential and the electric field strength inside the cylinder are equal:


 * $$ \varphi_i = - \frac { \rho_{0q}\rho^2 }{4\varepsilon_0 }+ \frac { \rho_{0q} }{2 \varepsilon_0 } \left[ \frac {L \sqrt {L^2+4 a^2} }{4} + a^2 \operatorname{arsinh} { \frac {L}{2a} } - \frac {L^2}{4} \right ], $$


 * $$ \mathbf E_i = \frac { \rho_{0q} \rho }{2\varepsilon_0 } \mathbf e_\rho, $$

where $$ L $$ is the length of the cylinder, $$ a $$ is the radius of the cylinder, $$ \mathbf e_\rho $$ is the unit vector along the cylindrical coordinate $$ \rho $$.

As it can be seen, the potential inside the cylinder depends on its length $$ L $$ logarithmically, due to the presence of the inverse hyperbolic sine $$ \operatorname {arsinh} {\frac {L} {2a}} $$. The internal electric field $$ \mathbf E_i $$ far from the ends of the cylinder at $$ z < < L / 2 $$ is directed perpendicular to the axis of rotation and is equal to zero on the axis of rotation, where $$ \rho = 0 $$.

The corresponding external electric potential and electric field strength outside the long cylinder are as follows:


 * $$ \varphi_o = - \frac { \rho_{0q} a^2 }{2\varepsilon_0 } \ln \frac {\rho}{a}  - \frac { \rho_{0q} a^2 }{4\varepsilon_0 }+ \frac { \rho_{0q} }{2 \varepsilon_0 } \left[ \frac {L \sqrt {L^2+4 a^2} }{4} + a^2 \operatorname{arsinh} { \frac {L}{2a} } - \frac {L^2}{4} \right ], $$


 * $$ \mathbf E_o = \frac { \rho_{0q} a^2 }{2\varepsilon_0 \rho } \mathbf e_\rho . $$

The above formulas require correction near the ends of the cylinder, since here the electric potential and field strength become functions not only of $$ \rho $$, but also of $$ z $$.

Long rotating cylinder
When the cylinder rotates with a constant angular velocity $$ \omega $$, the Lorentz factor of charged particles of matter becomes the function of $$ \rho $$:


 * $$ \gamma = \frac {1}{ \sqrt {1-v^2/c^2} }= \frac {1}{ \sqrt {1- \omega^2 \rho^2/c^2} } .$$

Taking this into account, the solution to Eq. (1) for the scalar potential, as well as for the field strength inside a rotating uniformly charged cylinder far from the ends of the cylinder will be as follows:


 * $$ \varphi_i = \frac { c^2 \rho_{0q} }{\varepsilon_0 \omega^2 } \left[ \sqrt {1- \omega^2 \rho^2/c^2} - \ln \left (1+\sqrt {1- \omega^2 \rho^2/c^2} \right ) -1 + \ln 2 \right ] + \frac { \rho_{0q} }{2 \varepsilon_0 } \left[ \frac {L \sqrt {L^2+4 a^2} }{4} + a^2 \operatorname{ arsinh } { \frac {L}{2a} } - \frac {L^2}{4} \right ], $$


 * $$ \mathbf E_i = \frac { \rho_{0q} \rho }{\varepsilon_0 \left (1+\sqrt {1- \omega^2 \rho^2/c^2} \right ) } \mathbf e_\rho . $$

Outside a long rotating cylinder, the scalar potential and electric field strength are expressed by the formulas:


 * $$ \varphi_o = -\frac { \rho_{0q} a^2 }{2\varepsilon_0 } \ln \frac {\rho}{a} + \frac { c^2 \rho_{0q} }{\varepsilon_0 \omega^2 } \left[ \sqrt {1- \omega^2 \rho^2/c^2} - \ln \left (1+\sqrt {1- \omega^2 \rho^2/c^2} \right ) -1 + \ln 2 \right ] + \frac { \rho_{0q} }{2 \varepsilon_0 } \left[ \frac {L \sqrt {L^2+4 a^2} }{4} + a^2 \operatorname{ arsinh } { \frac {L}{2a} } - \frac {L^2}{4} \right ], $$


 * $$ \mathbf E_o = \frac { \rho_{0q} a^2 }{2\varepsilon_0 \rho } \mathbf e_\rho . $$

Vector potential and magnetic field
The rotation of the charged matter of the cylinder leads to the appearance of the vector potential $$ \mathbf A $$ and the induction of the magnetic field $$ \mathbf B $$. For these values inside the cylinder, far from the ends of the cylinder, as a result of (2), the following is obtained:


 * $$ \mathbf A_i = \frac {\rho_{0q}}{\varepsilon_0}\left [ \frac {c^2 }{3 \omega^3 \rho} - \frac {c^2 }{3 \omega^3 \rho} \left( 1- \omega^2 \rho^2/c^2 \right)^{3/2} - \frac {\rho }{2\omega } + \frac {\omega \rho }{4 c^2} \left( L \sqrt { \frac {L^2}{4} +a^2 } - \frac {L^2}{2} \right) \right] \mathbf e_\phi, $$


 * $$ \mathbf B_i =\frac {\rho_{0q}}{\varepsilon_0}\left [ \frac {1 }{ \omega} \sqrt {1- \omega^2 \rho^2/c^2} - \frac { 1}{\omega } + \frac {\omega }{2 c^2} \left( L \sqrt { \frac {L^2}{4} +a^2 } - \frac {L^2}{2} \right) \right] \mathbf e_z, $$

where $$ \mathbf e_\phi $$ is a unit vector directed along the cylindrical coordinate $$ \phi $$, $$ \mathbf e_z $$ is a unit vector directed along the cylindrical coordinate $$ z $$. As it can be seen, the internal vector potential rotates around the axis of rotation of the cylinder. As for the magnetic field, it is directed along the rotation axis along which the $$ z $$ coordinate is measured. In this case, the magnetic field is maximum on the axis itself and tends to zero near the surface of the cylinder.

The external vector potential and the magnetic field of a long cylinder are determined by the formulas:


 * $$ \mathbf A_o = \frac {\rho_{0q}}{\varepsilon_0 \rho }\left [ \frac {c^2 }{3 \omega^3 } - \frac {c^2 }{3 \omega^3 } \left( 1- \omega^2 a^2/c^2 \right)^{3/2} - \frac {a^2 }{2\omega } + \frac {\omega a^2 }{4 c^2} \left( L \sqrt { \frac {L^2}{4} +a^2 } - \frac {L^2}{2} \right) \right] \mathbf e_\phi, $$


 * $$ \mathbf B_o = \nabla \times \mathbf A_o \approx 0 . $$

These formulas are reasonably accurate near the center of a long cylinder. However, as one approaches the ends of the cylinder, one should take into account the fact that significant additions appear in the formulas for the vector potential and magnetic field due to the dependence on the $$ z $$ coordinate. For an infinitely long cylinder, the above formulas can be used without restrictions.

Fedosin's theorem makes it possible to accurately calculate the magnetic field on the axis of rotation of charged rotating bodies. In particular, the magnetic field inside the cylinder depends on $$ z $$:


 * $$B_z(0 \le z \le L/2) = \frac {\mu_0 \omega \rho_{0q}}{2}\left [ \left (\frac {L}{2}+z \right ) \sqrt { \left ( \frac {L}{2} +z \right )^2 + a^2 } - \left ( \frac {L}{2} +z \right )^2 \right ] + \frac {\mu_0 \omega \rho_{0q}}{2}\left [ \left (\frac {L}{2} - z \right ) \sqrt { \left ( \frac {L}{2} - z \right )^2 + a^2  } - \left ( \frac {L}{2} - z \right )^2 \right ] . $$

At the center of the cylinder at $$ z = 0 $$, the magnetic field is:


 * $$B_z(z =0) = \frac {\mu_0 \omega \rho_{0q}}{2} \left ( L \sqrt { \frac {L^2}{4} +a^2} - \frac {L^2}{2} \right ) \approx \frac {\mu_0 \omega \rho_{0q} a^2}{2} . $$

If we take points on the axis of rotation outside the cylinder, then the magnetic field there looks like:


 * $$B_z(z \ge L/2) = \frac {\mu_0 \omega \rho_{0q}}{2}\left [ \left (z + \frac {L}{2} \right ) \sqrt {\left (z + \frac {L}{2} \right )^2 + a^2 } - \left (z - \frac {L}{2} \right ) \sqrt { \left (z - \frac {L}{2} \right )^2 + a^2  } + \left (z - \frac {L}{2} \right )^2 - \left (z + \frac {L}{2} \right )^2 \right ] \approx \frac {\mu_0 \omega \rho_{0q} a^4 L}{8 z^3}. $$

At the end of the cylinder at $$ z = L / 2 $$, we get


 * $$B_z(z = L/2) = \frac {\mu_0 \omega \rho_{0q}}{2} \left ( L \sqrt {L^2 + a^2} - L^2 \right ) \approx \frac {\mu_0 \omega \rho_{0q} a^2}{4}. $$

As a result, the magnetic field at the center is almost twice as large as at the end of the cylinder on the axis of rotation. This difference shows the degree of influence of edge effects and the need to take into account in (2) the dependence of the vector potential on the coordinate $$ z $$ near the ends of the cylinder.