HW-571-4-solution

HW-571-4-solution - 1 Phys 571 , Fall 2010 Solution to...

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Unformatted text preview: 1 Phys 571 , Fall 2010 Solution to Assignment #4. 1. The difference between the two gauges is A −A= Since 2. 3. (a) E = γ (E − V × B /c) − (γ − 1)V V·E V2 V·B V2 × ( f ) = 0, magnetic field is the same. − xyB 2 B = γ (B + V × E /c) − (γ − 1)V (b) There are infinitely many solutions to this problem. If in some reference frame K , which moves with velocity V, E B then in any other frame that moves parallel to K along the common direction of the fields E and B will be parallel (this follows from equations derived in (a)). Therefore, it sufficient to find a system K that moves perpendicularly to the plane B, E. Now, E × B = 0 since E B. It is convenient to write eqs of (a) for E and B (just replace E ↔ E , B ↔ B , V → −V) and use E × B = 0, V · B = V · E = 0 to obtain a quadratic equation for V/c. Its solution: E2 + B2 − V =E×B c Using the field invariants: 1 E 2 = [E 2 − B 2 + 2 1 B 2 = [B 2 − E 2 + 2 (E 2 − B 2 )2 + 4(B · E)2 ] (E 2 − B 2 )2 + 4(B · E)2 2(E × B)2 (E 2 − B 2 )2 + 4(B · E)2 ] (c) Using the field invariants we see that if E > B , there exists a frame in which B = 0 √ and E = E 2 − B 2 . Analogously, if E < B , there exists a frame in which E = 0 and √ B = B 2 − E 2. 2 E > B case: V=c E×B , E2 ˆ E =E E2 − B2 In any other frame moving along E with an arbitrary velocity magnetic field is absent. E < B case: V=c 4. B×E , B2 ˆ B = B B2 − E2 ...
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This note was uploaded on 02/24/2011 for the course PHYS 571 taught by Professor Krill during the Fall '10 term at Iowa State.

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HW-571-4-solution - 1 Phys 571 , Fall 2010 Solution to...

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