MadSci Network: Physics |
Your answer to the first question (the neutral wire case) is substantially correct and very insightful, as they are in fact the next points you raise! The other two questions are symmetrical, in the sense that from the point of view of the observer what you are looking at is always a wire made of charged particles (either the electrons or the protons), which looks either stationary or moving depending on the frame of reference you are sitting in: when you are moving along with the electrons the electrons look stationary, and the protons look moving backwards while when you move along with the protons it's the electrons which actually look moving along the wire direction. So the two situations are in fact identical from the point of view of electromagnetism and relativity (except of course for the fact that electrons and protons have opposite charges): bear in mind that there is no privileged frame of reference in the framework of relativity! The answer to your second question (what happens if there isn't any protons?) is that when you are in a frame where the wire is stationary you will not see any magnetic field, but you will see the electric field generated by the stationary distribution of charge. When you set the wire in motion, there will be a magnetic field generated, according to Ampere's law (http://en.wikipedia.org/wiki/Amperes_Law). And the effective current that you will see will grow with the speed of the wire, therefore also the magnetic field will! What is less immediate is that also the electric field generated by the wire in motion will slightly increase! The reason is the relativistic phenomenon known as "length contraction" (see either "Taylor and Wheeler" or http://en.wikipedia.org/wiki/Special_Relativity#Time_dilation_and_length_contraction) will increase the amount of charge you see per unit length i.e. the charge density of the wire! However, while the B field increases approximately linearly with the speed v of the wire, the E field actually grows proportionally to 1/Sqrt(1-v^2/c^2) [where c is the speed of light] and this growth can actually be disregarded for v<http://en.wikipedia.org/wiki/Special_Relativity#Time_dilation_and_length_contraction). In practice, if you initially had a certain amount of charge q per each segment of length l along the wire, when you move to M the same charge will be distributed on the same segment which however now appears to be of length l'=l/gamma, where gamma=1/sqrt(1-v^2/c^2). Since gamma>1, l' q/l=lambda: in M the wire appears as a wire with more charge per unit length!!! The electric field for the moving wire will be, from [1]: E=lambda gamma/(2 pi r epsilon0) The wire is moving though, so together with an electric field we will have a magnetic field! In order to derive the intensity of the field we have to figure out what the current is that corresponds to the moving wire... but this is straightforward given lambda' and v: I = lambda'*v=lambda*v*gamma So the magnetic field generated by the wire can be obtained right from Ampere's law: B= mu0 I / (2 pi r) = mu0 lambda gamma v/(2 pi r) You may have noticed already that in fact B and E are pretty similar: B = v E (epsilon0/mu0) And this relationship holds also when the wire is stationary, since in that case v=0 and therefore B=0 ! So we learned two things about this moving wire: - a magnetic field gets generated, which grows with v - the wire's electric field grows with respect to the stationary case Back to case 1 now! We can use what we just learned about a 'charged' wire to the 'neutral wire' case, using the so called "superposition principle". That is, nothing else than the fact that you can think your wires of protons and electrons as two separate problems, find the solution and then derive the solution of the combined problems as the sum of the fields from each of them. When you sit in the frame of reference where the protons are stationary and the electrons flow in the wire, you will have to add together the electric and magnetic fields of the protons and the electrons. When the wire is neutral and the current flows through it, the charge density of protons lambda is equal and opposite to the one for electrons (-lambda) therefore in presence of a current: - The protons will emit no magnetic field - the protons will emit a radial electric field which is E = lambda/(2 pi r epsilon0) - the electrons will emit a radial electric field which is E = -lambda / (2 pi r epsilon0) - the electrons will emit a tangential magnetic field of strength B = mu0 (-lambda) v/(2 pi r) So overall the system will emit a radial electric field E = 0 and a tangential magnetic field B = mu0 lambda v/(2 pi r). When we sit now in the frame of reference where the electrons look stationary: - The electrons will emit no magnetic field - the electrons will emit a radial electric field which is E = -lambda/(2 pi r epsilon0) - the protons will emit a radial electric field which is E = lambda / (2 pi r epsilon0) - the protons will emit a tangential magnetic field of strength B = mu0 lambda (-v)/(2 pi r) (note that the magnetic field is now generated by a positive charge density moving along the wire in a direction which is opposite to the one for the negative charge density in the case of the moving electrons, resulting in the same magnetic field strength and sign).
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