# Finding the induced EMF on a bar - Faraday¡s law

## Homework Statement

A conducting bar of length L moves with velocity v, in a rectangular region with a uniform and stationary magnetic field B_1. Near the bar, there is a cylindrical region of radius R with another magnetic field B_2 (it may vary on time). Both magentic field are normal to the plane of the picture, and point inwards. a)Find the expression of the induce EMF on the bar, indicating its polarity, and the potential difference between the top and botton of the bar.
b)Is it equal to the EMF induced in the circuit P1,P2,P3,P4?
c)What if the bar was non-conducting?

## Homework Equations

Faraday's law: $$\varepsilon=-\frac{d}{dt} \iint_{S(C)}\vec B \vec{ds}=-\iint_{S(C)}\frac{d \vec B}{dt} \vec{ds}+\oint_C \vec v \times \vec B \vec {dl}$$

## The Attempt at a Solution

a)Integrating over the red curve C and the surface enclosed by it, I get:
$$\varepsilon=-\iint_{S(C)}\frac{d \vec B}{dt} \vec{ds}+\oint_C \vec v \times \vec B \vec {dl}=\frac{dB_2}{dt}\pi R^2+vB_1L$$
I took positive EMF anti-clockwise, I may have messed up with the signs anyway.

Is this correct? I'm confused, shouldn't the EMF be the same whichever curve I take? (Note that if I take a curve such that it doesn't include the cylinder, the first term of the expression dissapears) My question is, how do I know which curve to take? It should be one which has all the points of the bar, but there are lots of them.

b)I would say definitely not, since there is no magnetic flux variation on that surface so EMF=0.

c)If the bar was non-conducting charges wouldn't move inside the bar, and so there would be no EMF there, but the potential difference would be the same?

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I tried further and now I seem to have found the EMF's induced by both the cylindrical region (because of its time-depending B) and the square one (because of the bar motion), not taking into account the other one. Assuming both results are right, how do I get the final solution, the 'total' EMF?

Taking only the cylindrical region and integrating over the blue curve: $$\phi=-\frac{B(t)\theta(t)R^2}{2}$$
where $$\theta=2\arctan{\frac{L}{2x(t)}}$$
then
$$\varepsilon=\oint \vec E \vec{dl}=- \frac{d\phi}{dt} = \frac{1}{2} \frac{dB}{dt} \frac{d\theta}{dt}R^2$$, and this is the EMF's produced by the B2 of the cylinder. Due to the symmetry, E is only tangential to the cylinder/circle.
The other EMF is the classic $$vBl$$

So if what I did is right, is the total EMF the sum of the other two?