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8.022pset10solution

8.022pset10solution - Massachusetts Institute of Technology...

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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 10: Displacement current, Maxwell’s equations, Radiation. TA: Yi Mao, [email protected] 1. Purcell 9.6: Radiation’s ~ E and ~ B in SI units. The source free Maxwell Equations read as follows. ~ ∇× ~ E =- ∂ ~ B ∂t (1) ~ ∇× ~ B = μ ² ∂ ~ E ∂t (2) ~ ∇· ~ E = (3) ~ ∇· ~ B = . (4) Consider the electromagnetic wave of the form ~ E = ˆ zE sin( y- vt ) ~ B = ˆ xB sin( y- vt ) . (5) Plug eqs.(5) into eqs.(1-4). Eqs.(3) and (4) are automatically satisfied. Eqs.(1) and (2) become ˆ xE cos( y- vt ) = ˆ xB v cos( y- vt )- ˆ zB cos( y- vt ) =- ˆ zμ ² E v cos( y- vt ) or, E = B v (6) B = μ ² E v. (7) From Eqs.(6) and (7) one can derive the value of speed v of electromagnetic wave in the vacuum: v = 1 / √ ² μ . (8) 2. Purcell 9.8: Wave in a box. For the EM field ~ E = ˆ zE cos kx cos ky cos ωt (9) ~ B = B (ˆ x cos kx sin ky- ˆ y sin kx cos ky ) sin ωt, (10) the two divergencelessness equations of Purcell eqs.(16) ~ ∇· ~ E = 0 ; ~ ∇· ~ B = 0 , can be easily verified. The other two equations give ~ ∇× ~ E + 1 c ∂ ~ B ∂t = ( ωB c- E k )(ˆ x cos kx sin ky- ˆ y sin kx cos ky ) cos ωt = 0 ~ ∇× ~ B- 1 c ∂ ~ E ∂t = ( E ω c- 2 B k )ˆ z cos kx cos ky sin ωt = 0 or E = ω kc B (11) E = 2 kc ω B . (12) So the condition is that ω = √ 2 ck (13) E = √ 2 B . (14) The magnetic field inside a square metal box of size π/k is shown in Figure 1. Note that the box must run from- π/ (2 k ) to π/ (2 k ) in both x and y in order to satisfy the...
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8.022pset10solution - Massachusetts Institute of Technology...

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