A metal bar moves upwards near a long current carrying wire....

[jisbon]

Homework Statement

Bar moves upwards near a long current carrying wire. Current = 50A. What is the p.d between A and B?

Relevant Equations

$B =\frac{\mu I}{2\pi r}$

Not sure what to do here, but I do know that magnetic field is pointing in the paper according to the right hand grip from current carrying wire.
Hence I can calculate magnetic field to be:
$B = \mu_{0} I/2\pi r=(4\pi * 10^{-7})(50)/2\pi (0.004) = 3.94784*10^{-7}T$

Now from what I understand, there are free electrons inside the conducting bar. The positive charges move to left while negative charges move to right. So for a free electron to remain in bar,

$qE=bqv$
$E = vb$
I also know p.d = Efield in the bar * Length of bar
Which means$p.d = vBL$
However, subbing in B from above does not give me the answer.
Can anyone explain the concepts involved in this question and what I might have done wrong?

 

[Delta2]

I don't see any mistake in your reasoning (though it is a bit simplified, it still holds in this case because the velocity and the magnetic field do not vary along the length of the bar) so I expect the motional EMF to be $$\text{EMF}=\int_0^L(\vec{v}\times \vec{B})\cdot \vec{dl}=BvL$$. Perhaps you did a mistake in the arithmetic. What is the answer provided by the book?

There is definitely a mistake in the arithmetic for B, I get $B=2.5\times 10^{-3}T$

 

[jisbon]

Welp an arithmetic error :/
Is there a more detailed reasoning behind this concept though? Thanks

I don't see any mistake in your reasoning (though it is a bit simplified, it still holds in this case because the velocity and the magnetic field do not vary along the length of the bar) so I expect the motional EMF to be $$\text{EMF}=\int_0^L(\vec{v}\times \vec{B})\cdot \vec{dl}=BvL$$. Perhaps you did a mistake in the arithmetic. What is the answer provided by the book?

There is definitely a mistake in the arithmetic for B, I get $B=2.5\times 10^{-3}T$

 

[Delta2]

Welp an arithmetic error :/
Is there a more detailed reasoning behind this concept though? Thanks

Read Wikipedia article on Faraday's law and especially the proof section
https://en.wikipedia.org/wiki/Faraday's_law_of_induction#Proof
Other than that your reasoning in the OP is valid only if the magnetic field is uniform across the wire loop we are integrating (in this case the magnetic field is not uniform but across the bar length is uniform, if the bar was oriented perpendicular to the wire with its velocity parallel to the wire the EMF would be different than BuL). In the most generic case that the magnetic field and/or velocity varies through out the points of the wire loop we have to calculate the integral

$$\int_{Loop}(\vec{v}\times \vec{B})\cdot d\vec{l}$$

Try to calculate this integral in the case of a bar that is perpendicular to the wire with its velocity parallel to the wire. You ll se that it is quite different than $BuL$ (if you do it correctly you ll have $\ln{L+\alpha}$ terms)/