hw9solns - PHYSICS 273 HW #9 SOLUTIONS 9-20 Let the...

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PHYSICS 273 HW #9 SOLUTIONS
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9-20 Let the electric field be   0 sin . E kx t Its energy density is   2 2 2 0 0 0 0 11 sin with average . 24 E kx t E The kinetic energy density of electrons can be found from   2 22 2 2 2 2 2 0 0 0 0 2 2 2 1 1 1 1 cos with average 2 2 4 4 p ee ne ne nmv E kx t E E mm  (Note that the electron velocity is     00 sin cos .) t eE eE v kx t dt kx t      The magnetic field can be found from     sin / E B E kx t k    , which yields the magnetic energy density     2 2 2 2 2 0 1 1 1 1 sin 2 2 2 p B E E kx t       . Its average is therefore 2 2 1 1 4 p E And the total energy density is then 0 0 0 0 1 1 (rms) 42 pp EE    This indicates that in electromagnetic waves in a plasma, energy equipartition holds between the electric field energy density and the sum of electron kinetic energy density and magnetic energy density.
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This note was uploaded on 12/28/2011 for the course PHYSICS 273 taught by Professor Monroec during the Fall '11 term at Maryland.

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hw9solns - PHYSICS 273 HW #9 SOLUTIONS 9-20 Let the...

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