Hall Effect - Properties of a slab of metal

[smahapatra3]

Homework Statement

A slab made of unknown material is connected to a power supply as shown in the figure. There is a uniform magnetic field of 0.5 tesla pointing upward throughout this region (perpendicular to the horizontal slab). Two voltmeters are connected to the slab and read steady voltages as shown. (Remember that a voltmeter reads a positive number if its positive lead is connected to the higher potential location.) The connections across the slab are carefully placed directly across from each other. The distance w = 0.15 m. Assume that there is only one kind of mobile charges in this material, but we don't know whether they are positive or negative.

In the steady state, the current moves straight along the bar, so the net sideways force on a moving charge must be zero. Use this fact to determine the drift speed of the mobile charges

Homework Equations

F=q(E+vXB)
Eperp=uEparallelB

The Attempt at a Solution

It's in steady state so I know the electric and magnetic forces are equal. So I think the equation is Eperp=vB. I tried to find E using volts/length. And then divide by B but that was wrong.

 

[Steve4Physics]

This is an old (11+ years at time of answering) question. With no possibility of interacting with the OP, I've decided to provide a fairly detailed answer to help anyone encountering the thread.

A slab made of unknown material

Note: the title says ‘metal’ not ‘unknown material’. But since we are told sign of the charge-carriers is unknown, the title is incorrect.

It's in steady state so I know the electric and magnetic forces are equal. So I think the equation is Eperp=vB. I tried to find E using volts/length. And then divide by B but that was wrong.

That's the right aproach. But the OP hasn't shown any working, so we can't tell what's going wrong. Normally the OP would now be requested to provide the working in full.
___________________________________________________________________________________

Here's an answer...

We don't actually need the sign of the charge-carrier unless we want the direction of drift. But, if required, we can easily find the sign...

Call the front and back parts of the block F and B respectively.

From the diagram and the 0.73V meter-reading, we know there are either:
a) positive charge carriers moving from B to F or
b) negative charge carriers moving from F to B.

Applying Fleming’s left hand rule, we can tell to which side the charge-carriers are pushed. Either:
a) positive charge carriers (moving from B to F) would be pushed to the left side of the slab; this would create a charge-build-up which makes the left side positive; or
b) negative charge carriers (moving from F to B) would also be pushed to the left side of the slab; this would create a charge-build-up which makes the left side negative.

From the -0.00027V meter-reading we can tell the left side is positive, so case a) applies: the charge-carriers are positive.

In the left/right direction, the magnitude of the magnetic force on a charge-carrier is qvB (where v is the required drift velocity in the direction from B to F). This force's direction is right-to-left.

In the left/right direction, the magnetic force on a charge carrier is balanced by electric force, magnitude qE, arising from the asymmetric left/right carrier-distribution This force's direction is left-to-right

$qvB = qE$
$v = \frac E B$

Note it is the 'sideways' electric field we are interested in. This electric field has magnitude $|E| = \frac {|V|}{d} = \frac {0.00027V}{0.08m} = 0.003375V/m$

The velocity follows from the above. Since B is only given to one sig. fig. the velocity should be correspondingly rounded.

Note, not all the data supplied in the question are used. For example the value of 'w' is not required.

EDIT: typo's korekted