Explore BrainMass

Elctromagnetic induction and the Lorentz force

1. An infinitely long, straight wire carries a constant current I along the z-axis. A rectangular loop of conducting wire (dimensions = a x b) is located in a radical plane at a distance x from the wire, as shown.

(see the attached file for the diagram)

(a) Find the flux of the magnetic-field B through the loop.
(b) If the loop moves in and out in the radial direction such that x(t) is a given function of time, determine the induced voltage V(t).
(c) Show that applying the Lorentz Force Law, F = q(E + V x B), yields the same result as obtained in part (b) based on Faraday's Law.
(d) What happens to the voltage V(t) if x is kept constant, but the loop is moved up and down parallel to the z-axis? Justify your answer by using both Faraday's Law and Loretz's Law.


Solution Preview

The magnetic field at distance r from the wire is:

B(r) = mu_0/(2 pi r) I

B(r) points into the paper if the current moves upward.

The flux through the loop is:

phi = mu_0/(2 pi ) I b Integral over r from x to x + a of 1/r dr =

mu_0/(2 pi ) I b Log[1+a/x]

Here we count a flux into the paper as positive. This means that according to the right hand rule we must assign to the loop a clockwise orientation.

The induced voltage (a.k.a. emf) is given by:

V = - d phi/dt = mu_0/(2 pi ) a b/(x^2 + a x) dx/dt I

V calculated this way is always equal to the line integral of the induced electric field along the loop. Here we assume that the loop is closed, let's assume that the plus and minus terminals shown in the figure are connected by a voltmeter. This line integral has to be evaluated in the orientation consistent with the way the flux is defined. In our case this orientation is clockwise. The voltmeter will measure the potential difference between the plus terminal and the minus terminal, which is given by the line integral of E dot ds from the plus to the minus terminal (along the wire which runs through the voltmeter, we assume that the voltmeter is placed in between the plus and minus terminal so that the computation of the flux above is correct). Now if we add to this the line integral of E dot ds from the ...

Solution Summary

The following posting helps with problems involving electromagnetic induction and the Lorentz force.