# Induced potential in a bar due to a magnetic field

1. Jun 9, 2009

### Faloren

1. The problem statement, all variables and given/known data

http://img268.imageshack.us/img268/6536/physics1.png [Broken]

2. Relevant equations

Lorentz Force = $F = q(E+(v\times B))$

3. The attempt at a solution

i. $I = \frac{dQ}{dt}$

So, dQ = number of electrons x area dx

i.e. $I = \frac{n e d^2}{dt}dx = n e d^2 v_d$

ii

Presumably the velocity of the electrons combined with the magnetic field means that we use the Lorentz force - giving us a force on each electron. The electrons move to one side and cause the bar to effectively polarise. The electrons will stop moving when the electrostatic attraction between the negatively charged region of the bar and the postively charged region of the bar are equal.

So:

$Force = Eq$ and $Force = q(v \times B) = qvB$

Combining the two gives:

$Eq = qvB, E = vB = \frac{I}{ned^2}|B|$

Electric field can be used to work out the potential:

$V_b - V_a = \int{E.dl} = V = Ed = \frac{I}{ned}|B|$

The force is to the left of the diagram, using the right hand rule and so the electric field points from right to left (positive charges on the right hand side, so the field goes to the negative region - where all the electrons are).

iii. By observing which way the electric field is - or rather what the potential difference is between one side of the bar and the other, it should tell us what the charge carriers are. If they're negative we will see a potential as above, if they're positive it will be as above, but negative.

Does that look ok?

(i haven't drawn a diagram, but as i said, i would expect the field to flow from right to left, positive to negative)

Last edited by a moderator: May 4, 2017