Question

The conducting slab has a thickness b and a width a. The magnetic field is constant and points out of the paper. The voltmeter is disconnected from the circuit. There is a steady flow of a horizontal current i moving from left to right, or a steady flow of electrons from right to left with an average drift velocity vd. In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force?

1) e = vd i b
2) e = i b
3) e = a vd b b
4) e = v² d b
5) e = vd b
6) e = a vdb
7) e = 3 vd b
8) e = vd² b

Answer 1

The correct magnitude of the electric field that cancels out the magnetic force on electrons due to the Hall effect in the given scenario is E = vd B, where vd is the average drift velocity and B is the magnetic field strength.

Step-by-step explanation:

The question relates to the Hall effect observed when a current-carrying conductor is placed in a magnetic field. When a steady current i flows through a conductor of width a and thickness b, with a magnetic field pointing out of the page, the electrons moving with average drift velocity vd experience a magnetic force. This force is countered by an electric field E, and the magnitude of this electric field can be found using the relationship E = vdBi.

Using the Lorentz force law, F = qvdB for magnetic force on a charge and F = qE for electric force, we can equate the two to cancel the magnetic force with electric force as qE = qvdB. Simplifying, we find E = vdB. To find the relationship involving current, we use I = nevdA, and since A = ab (cross-sectional area of the conductor), we substitute to get E = IB/neb. Since we are looking for an expression that involves the average drift velocity and not electron density or charge, the correct magnitude of the electric field that produces this force is E = vdB.