571-solution

571-solution - 1 Phys 571 , Fall 2010 FINAL EXAM Solutions...

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1 Phys 571 , Fall 2010 FINAL EXAM Solutions 1. (a) Fields outside the cylinder are (frame K ) E = 2 λ r ˆ r , B = 2 cr I ˆ φ , r a and inside E = 2 λr a 2 ˆ r , B = 2 r ca 2 I ˆ φ , r a, where a is the cylinder radius. Frame K 0 must move along the z -axes. Indeed, in that case E k = B k = 0 in both frames. Now E 0 = γ ( E + 1 c V × B ) , B 0 = γ ( B - 1 c V × E ) From B 0 = 0 it follows that V = ˆ E × ˆ B cB E = ˆ r × ˆ φ I λ = ˆ z I λ The same result can be obtained more easily using the velocity transformation and requiring that in K 0 all charges are static v 0 = 0. In that case, from j = vq/ ( AL ), where A is the cross sectional area and L is length of the cylinder, it follows that V = v = jAL/q = I/λ . (b) E 0 = 2 λ r ˆ r r 1 - I 2 λ 2 c 2 (c) When I cλ K 0 does not exist. 2. (a) Positions of the three image charges are shown in the figure below. Total potential is
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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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571-solution - 1 Phys 571 , Fall 2010 FINAL EXAM Solutions...

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