Lecture-2 Sept 17, 2010 Motional emf

Today we continued with E-B fields of a point charge q moving with a uniform velocity v. We found that (1) The E-field at a point at an instant is radial from the instantaneous position of the charge. So if you wish to get the field at P at time t, and the charge is at A at this time t, the field at t is along AP (q positive). (2) The magnitude of the E-field is largest in the direction perpendicular to the direction of motion and smallest in the direction of motion.


Then we started Motional emf. Questions: What is emf?, What is driving force and what is maintaining force? What charge distribution creates electric field in a wire connected to a battery, resistances etc? The most juicy part was the calculation of magnetic field and electric field inside a conducting rod moving with a uniform velocity v in a uniform magnetic field. The rod, velocity and the B-field were taken mutually perpendicular.
Why should there be an E-field inside the rod, where I had only applied B-field? This is because the magnetic force pushes free electrons on one side and there is a charge distribution coming up on the surface of the moving conductor.


This charge distribution also moves with the rod and hence produces a B-field other than the applied B-field. Thus the B-field inside the rod (and in the vicinity of the rod) will the different from the applied field B. The task is to get these fields. And to do this we go to the rod frame which is S'. Clearly distinguish applied fields in S, Applied Fields in S', Inside (rod) Fields in S and Inside fields in S'. At this moment we know Applied fields in S, and wish to get inside Fields in S. Using Direct transformation equations, get the Applied Fields in S'. Though the Applied E-field is zero in S, it is not zero in S'. It is in y-direction in the scheme taken in class. The Applied B field has somewhat larger magnitude in S' though the direction is the same.


Now that we know the applied fields in S', we turn our attention to fields inside the rod in S'. As the rod is at rest and there is an E-field in y-direction, charges will redistribute on the surface of the rod and finally the E-field inside the rod will be zero. Remember we are talking from S' frame. The redistributed charges are at rest and so produce no magnetic field. The magnetic field inside the rod is THEREFORE equal to the magnetic field Applied (all talks in S'). Hence we obtain E and B fields inside the rod in S'.


Since we know E and B fields inside the rod in S', we do Back transformation to get E and B fields inside the rod in S. And that completes the task.


What do we find for the fields? Now I am talking from S. The job of S' is done. All our experiments are in S. We took help of S' only to get the fields inside the rod in S. As the surface charge distribution on the rod was not known, we could not have directly calculated fields inside the rod in S.
1) The B-field inside the rod is larger than the applied B-field by the factor 1/(1-v²/c²).
2) There is an E-field inside the rod. You can check, E = - v x B as expected in steady state.


Hope it had been a challenging session. I was benefited by the 30-m questions from various students after the class. Thank you.