Abstract
Energy and momentum are exchanged between particles and field when the electromagnetic field accelerates the charged particles and the particles radiate electromagnetic waves. In isolated systems with charges and electromagnetic field, energy and momentum are conserved.
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- 1.
We use cylindrical coordinates.
- 2.
This is correct for any component with a current I flowing inside and a voltage drop V.
- 3.
We get the same result if we consider the discharge of the condenser with a resistor R. The charge Q as a function of the time is:
$$ Q=Q_0 e^{-{t \over \tau }} $$where \(\tau =RC\), and the electric field can be written:
$$ \quad E={\sigma \over \epsilon _0}={1 \over \epsilon _0}{Q_0 \over {\pi r^2 }}e^{-{t \over \tau }}\,. $$The Poynting’s vector is:
$$ S={1 \over \tau }\left( {Q_0 \over {\pi r^2}}\right) ^2{r\over {2 \epsilon _0}}e^{-{{2t} \over \tau }} $$and its integral over the lateral surface during the discharge gives the electrostatic energy initially in the condenser as in (9.9).
- 4.
A. Sommerfeld, Electrodynamics, (1952), Academic Press, New York, pp. 125–130.
- 5.
See Problem 9.2 of this chapter.
- 6.
I. Galili and E. Goihbarg, Energy transfer in electrical circuits: a qualitative account, American Journal of Physics 73, pp. 141–144, (2005).
- 7.
Other generators are based on thermal, luminous or mechanical phenomena but the electromotive field description is still valid.
- 8.
See for instance: F.S. Crawford Jr, Waves, Berkeley Physics Course, Vol. 3, (1965), McGraw-Hill, Sect. 7.4.
- 9.
For more details see for instance: J.D. Jackson, Classical Electrodynamics, cited, Sect. 12.10; L.D. Landau-E. M. Lifšits, The Classical Theory of Fields, Chapter IV, Sect. 33.
- 10.
The force can be also calculated from the principle of virtual work.
- 11.
This formula can be found from the principle of virtual work. In a electrical network with a generator, a solenoid and a resistor, the sum of the work \(\delta L_{ext}\) from an external force and of the work \(\delta L_{gen}=fi dt\) of the generator of electromotive force f, which keeps constant the current i, has to be equal to the sum of the changes of the magnetic energy \(\delta U_m\) and of the energy \(\delta W_R=i^2 R dt\) dissipated in the resistor. We can write:
$$ \delta L_{ext}+ \delta L_{gen}=\delta U_m + \delta W_R \quad \quad \text {where:} \quad \quad U_m =\frac{1}{2}Li^2. $$At the same time the equation of the circuit is:
$$ f-\frac{d \varPhi (B)}{dt} =Ri \quad \quad \varPhi (B)=Li. $$We find:
$$ \delta L_{ext}+id(Li)+i^2 Rdt=\delta \left( \frac{1}{2}Li^2\right) +i^2R dt $$and from \(\mathbf {F}_{magn}=- \mathbf {F}_{ext}\)
$$ \delta L_{ext}=\mathbf {F}_{ext} \cdot \mathbf {ds}=-\mathbf {F}_{magn} \cdot \mathbf {ds}=-\delta U_m \quad \rightarrow \quad F_{magn}=\left[ \frac{dU_m}{ds}\right] _{i=const}. $$
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Lacava, F. (2016). Energy and Momentum of the Electromagnetic Field. In: Classical Electrodynamics. Undergraduate Lecture Notes in Physics. Springer, Cham. https://doi.org/10.1007/978-3-319-39474-9_9
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DOI: https://doi.org/10.1007/978-3-319-39474-9_9
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