# Crossed Fields The Hall Effect

Crossed Fields  The Hall Effect

As we just discussed, a beam of electrons in a vacuum can be deflected by a magnetic field. Can the drifting  conduction electrons in a copper wire also be deflected by  a magnetic field? In 1879, Edwin H. Hall, then a 24-year-old graduate student at  the Johns Hopkins University. showed that they can. This Hall effect allows us to find out whether the charge carriers ina conductor are positively or negatively charged. Beyond that, we can measure the number of such carriers per unit volume of the conductor, Figure 29-8a shows a copper strip of width d, carrying a current i whose conventional direction is from the top of the figure to the bottom. The charge carriers are electrons and, as we know, they drift (with drift speed Vd) in the opposite direction, from bottom to top. At the instant shown in Fig. 29-8a, an external magnetic  field ii, pointing into the plane of the figure, has just been turned on. From Eq.29-2 we see that a magnetic deflecting force F8 will act on each drifting electron, pushing it toward the’ right edge of the strip. As time goes on, electrons move to the right, mostly piling up on the right edge of the strip, leaving uncompensated positive charges in fixed positions at the left edge. The separation of positive and negative charges produces an electric field £  within the strip, pointing from left to right in Fig. 29-8b. This field exerts an electric force FE on each ‘electron, te ding to push it to the left. . An equilibrium quickly develops in which the electric force on each electron  builds up until it just cancels the magnetic force. When this happens, as Fig. 29-8bshows, the force due to ii and the force due to ‘£ are in balance. The drifting electrons then move along the strip toward the top of the page at velocity Vd’ with no further collection of electrons on the right edge of the strip and thus no further increase in  the electric field £.A Hall potential difference V is associated with the electric field across strip width d. From Eq. 25- 2, the magnitude of that potential difference/is

V = Ed

By connecting a voltmeter across the width, we can measure the potential difference between the two edges of the strip. Moreover, the voltmeter can tell us which edge is at higher potential. For the situation of Fig. 29-8a, we would find that the left .

edge is at higher potential, which is consistent with our assumption that the charge carriers are negatively charged.
For a moment, let us make the opposite assumption, that the charge carriers in current i are positively charged (Fig. 29-8c). Convince yourself that as these charge carriers move from top to bottom in the strip, they are pushed to the right edge by 18 and thus that the right edge is at higher potential. Because that last statement is contradicted by our voltmeter reading, the charge carriers must be negatively charged. Now for the quantitative part. When the electric and malign tic forces are in balance  and 29-3 give us

in which I (= A/d) is the thickness of the strip. With this equation we can find n from measurable quantities. It is also possible to use the Hall effect to measure directly the drift speed Vd o  the charge carriers, which you may recall is of the order of centimeters per hour. In this clever experiment, the metal strip is moved mechanically  through the magnetic f eld in a direction opposite that of the drift velocity of the charge carriers. The speed of the moving strip is hen adjusted until the Hall potential difference vanishes. At this condition, with no Hall effect, the velocity of the charge carriers with respect to the laboratory frame must be zero, so the velocity of the strip must be equal in magnitude but opposite the direction of the velocity of the negative charge carriers.

Sample Problem

Figure 29-9 shows a solid metal cube, of edge length d = 1.5 ern, moving in the positive y direction at a constant velocity v of magnitude 4.0 m/s. The cube moves through a uniform magnetic field if of magnitude 0.050 T directed toward positive z.

(a) Which cube face is at a lower electric potential and which is at a higher electric potential because of the motion through the field?

SOLUTION: One Key Idea here is that, because the cube is moving through a magnetic field E, a magnetic force 1B acts on its charged  articles. including its conduction electrons. A second Key Idea is how FB causes an electric potential difference between certain faces of the cube. When the lube first begins to move through the magnetic field. its electrons do also. Because each electron has charge q and is moving through a magnetic field with velocity ii, the magnetic force 1B acting on it is given by Eq. 29-2. Because q is.

Most of the electrons are fixed in place in the molecules of the cube. However, because the cube is a metal, it contains conduction electrons that are free to move. Some of those conduction electrons are deflected by FB to the left cube face, making that face  negatively charged and leaving the right face positively charged.This charge separation produces an electric field E directed from  the positively charged right face to the negatively charged left face.Thus, the left face is at a lower electric potential  and the right face is at a higher electric potential.

(b) What is the potential difference between the faces of higher and lower electric potential?

SOLUTION: The Key Ideas here are these: The Key Ideas here are these:

2. When the cube had just begun to move through the magnetic When the cube had just begun to move through the magnetic field and the charge separation had just begun, the magnitude  of E began to increase from zero. Thus, the magnitude of False began to increase from zero and was initially smaller than the magnitude FB• During this early stage, the net force on any election was dominated by FB’ which continuously moved additional electrons to the left cube face, increasing the charge separation.

3. However, as the charge separation increased, eventually magnitude FE became equal to magnitude FB• The net force on any  election was then zero, and no additional electrons were moved to the left cube face. Thus, the magnitude of FE could nOl increase further, and the electrons were then in equilibrium.

We seek the potential difference V between the left and right cube faces after equilibrium was reached (which occurred quickly).

We can obtain V with Eq. 29-9 (V = Ed) provided we first find the magnitude E of the electric field at equilibrium. We can do so with the equation for the balance of forces (FE = F B)’ For FE’ we substitute IqIE. For FB’ we substitute IqlvB sin cf>  from Eq. 29-3. From Fig. 29-9, we see that the angle cf> between vectors it and 8is 90°; so sin ‘” = I and FE = FB yields IqlE = IqlvB sin 90° = IqlvB. This gives us E = vB, so Eq. 29-9 (V = Ed) becomes V = vBd. (29-13) Substituting known values gives us V = (4.0 m/s)(0.050 T)(0.015 m) = 0.0030 V = 3.0 mY.