IL Fields and sources transformation. A long straight wire of radius R carries u
ID: 1836896 • Letter: I
Question
IL Fields and sources transformation. A long straight wire of radius R carries uniform current I and zero charge density, according to an observer in the wire's rest frame (the un rimed frame). This current is due to electrons moving at relativistic velocity u and positive ions at rest. Another observer moves at relativistic velocity v parallel to the direction of the wire (the primed frame) at radius r from the wire axis. a. (6 pts.) What's the electromagnetic fields seen by the moving (primed) observer? b. (6 pts.) What's the charge density seen by the moving (primed) observer? e. (7 pts) What are the velocities of the electrons and ions according to the moving (primed) observer? d. (6 pts.) Explain why one observer sees a charge density, but the other does not.Explanation / Answer
The electrons are moving at a speed of u an the +ve ions are at rest in frame S which is the rest frame.
consider a length l of the conuctor then
l = lu, the crrent is due to the flosing electrons nd the flow of charge per unit time is lu
let us consder the frame S' which is mvoing at aspeed of v to the right
The speed of the +ve ions as seen in the frame S' is
v+ = -v
speed of -ve electrons as seen in S'
v- = u+v/(1+uv/c2)
As there is difference in releative speeds of the line charge in S' the length of the charge of _ve and +ve charges is different in S', where as in S the +ve and -ve charges are equally spaced and the net charge is 0.
The net charge of the length l of the conductor is
= -luv/c2(1-v2/c2)1/2 = -Iv/c2(1-v2/c2)1/2