Question

Question: the Hall effect can be used to determine the sign of the mobile charges in a particular conducting material. A bar of a new kind of conducting material is connected to a battery as shown in Figure 20.85. In this diagram, the x-axis runs to the right, the y-axis runs up, and the z-axis runs out of the page, toward you. A voltmeter is connected across the bar as shown, with the leads placed directly opposite each other along a vertical line. In order to answer the following question, you should draw a careful diagram of the situation, including all relevant charges, electric fields, magnetic fields, and velocities.

Initially, there is no magnitude filed in the region of the bar. (a) Inside the bar, what is the direction of the electric field $\overrightarrow{\mathbf{E}}$due to the charges on the batteries and the surface of the wires and the bar? This is the electric field that drives the current in the bar. (b) If the mobile charges in the bar are positive in what direction do they move when the current runs? (c) If the mobile charges in the bar are negative, in what direction do they move when the current runs? (d) In this situation (zero magnetic fields), what is the sign of the reading on the voltmeter?

Next, large coils (not shown) are moved near the bar. And current runs through the coils, making a magnetic field in the -z direction (into the page). (e) If the mobile charges in the bar are negative, what is the direction of the magnetic force on the mobile charge? (f) If the mobile charges in the bar are negative, which of the following things will happen? (1) Positive charge will accumulate on the top of the bar. (2) The bar will not becomes polarized. (3) Negative charge will accumulate on the left end of the bar. (4) Negative charge will accumulate on the top of the bar. (g) If the mobile charges in the bar are positive, what is the direction of the magnetic force on the mobile charges? (h) If the mobile charges in the bar are positive, which of these things will happen? (1) positive charge will accumulate on the top of the bar. (2) The bar will not becomes polarized. (3) Positive charge will accumulate on the right end of the bar. (4) Negative charge will accumulate on the top of the bar.

You look at the voltmeter and find that the reading on the meter is $\mathbf{-}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{\text{volts}}$. (i) What can you conclude from this observation? (Remember that a voltmeter gives a positive reading if the positive lead is attached to the higher potential location.) (1) There is not enough information to figure out the sign of the mobile charges. (2) The mobile charges are negative. (3) The mobile charges are positive.

Short answer

(a) The direction of the electric field is $x$.

(b) If mobile charges are positive, they will move in the direction of the electric field which drives the current in the bar that is along $x$direction.

(c) If mobile charges are negative, they will move opposite to the electric field which drives the current in the bar that is along the $-x$direction.

(d) The sign of the voltmeter is negative.

(e) The direction of the magnetic force on the mobile charges is $y$.

(f) Option 1 is the correct answer. The positive charge will accumulate on the top of the bar.

(g) The direction of the magnetic force on the mobile charges is $y$.

(h) Negative charge will accumulate on the top of the bar. Thus, option (4) is the correct answer.

(i) The mobile charges are negative. Thus, option (2) is the correct answer.

Step-by-step solution

A concept:

Hall Effect is the production of a voltage difference across the electrical conductor, transverse to a current in the conductor and to an applied magnetic field perpendicular to the transverse current.

An electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on the positive test charge. The electric field is radially outward from the positive charge and radially inward toward the negative point charge.

Sketch:

The below sketch illustrates the experimental setup for a Hall Effect.

(a) The direction of the electric field   due to the charges on the batteries and the surface of the wires and the bar:

Electrons in the bar are attracted towards the positive terminal of the battery. So, electrons will flow from the right to the left of the bar. The direction of the electric field $\left(E\right)$that drives the current in the bar is always opposite to the flow of electrons, i.e. from left to right (or) along x direction.

Hence, the direction of the electric field is $x$.

(b) Direction do they move when the current runs:

If mobile charges are positive, they will move in the direction of the electric field $\left(E\right)$ which drives the current in the bar that is along $x$ direction.

(c) If the mobile charges in the bar are negative, in what direction do they move when the current runs:

If mobile charges are negative, they will move opposite to the electric field $\left(E\right)$ which drives the current in the bar that is along the $-x$ direction.

(d) The sign of the reading on the voltmeter:

In the absence of a magnetic field (zero magnetic field), there is no magnetic force on the mobile charges to create a transverse electric field across the vertical direction of a bar (due to the Hall Effect). Since the transverse electric field is not present in the bar for this situation, the voltmeter shows a zero reading in it.

Hence, the sign of the voltmeter is negative.

(e) The sign of the reading on the voltmeter?

If mobile charges are negative, they will move opposite to the electric field $\left(E\right)$. Therefore the direction of velocity $\left(\overrightarrow{v}\right)$ of the negatively mobile charges is in $-x$direction.

It is given that, the magnetic field $\left(\overrightarrow{B}\right)$pointing along the $-z$direction (into the page).

The magnetic force acting on the negative mobile charges is obtained by taking general formula,

$\begin{array}{rcl}\overrightarrow{F}& =& -q\left(\overrightarrow{v}\times \overrightarrow{B}\right)\\ & =& -q\left[v\left(-\hat{x}\right)\times B\left(-\hat{z}\right)\right]\\ & =& -qvB\left(-\hat{y}\right)\\ & =& qvB\left(\hat{y}\right)\end{array}$

Hence, the direction of the magnetic force on the mobile charges is $y$.

(f) If the mobile charges in the bar are negative:

As you know, the direction of the magnetic force on the mobile charges is $y$. Due to this force, a positive charge will accumulate on the top of the bar and negative charges will accumulate on the bottom of the bar.

This creates a transverse electric field across the vertical direction of a bar due to Hall Effect which points from a positive charge to a negative charge i.e., along the downward $-y$ direction.

Hence, the correct option is (1). The positive charge will accumulate on the top of the bar.

(g) If the mobile charges in the bar are positive:

It is given that, the magnetic field $\left(\overrightarrow{B}\right)$pointing along the $-z$direction (into the page).

The magnetic force acting on the negative mobile charges is obtained by using general formula,

localid="1662213242991" $\begin{array}{rcl}\overrightarrow{F}& =& q\left(\overrightarrow{v}\times \overrightarrow{B}\right)\\ & =& q\left[v\left(\hat{x}\right)\times B\left(-\hat{z}\right)\right]\\ & =& -qvB\left(\hat{x}\times \hat{z}\right)\\ & =& -qvB\left(-\hat{y}\right)\end{array}$

localid="1662213343962" $\overrightarrow{F}=qvB\left(\hat{y}\right)$

Hence, the direction of the magnetic force on the mobile charges is $y$.

(h) If the mobile charges in the bar are positive:

If mobile charges are positive, the accumulation of the charges on the top and bottom of the bar is just reversed as in part of (f).

Hence, the accumulation of the charge on the top of the bar is a negative charge. Hence, the correct option is (4).

(i) Concussion of the given observation:

This sets a transverse electric field pointing from the top to the bottom of a bar (downward), so this makes the reading of the voltmeter would be negative, as is observed.

Therefore, the conclusion is that the charge carriers are negative.

Hence, the correct option is (2).