Conducting rod moving in magnetic field on a railing problem

Question

We have a conducting rod (lenght $l$) moving with constant velocity $v$, on two symmetrical railings, and and everything is conducting. Rod has some resistivity $\rho$, while railings are purely conducting. Magnetic field applied is perpendicular and constant. We need to find electric field vector inside the rod.

Now I applied faraday's law of magnetism: $$V=-\frac{d\phi}{dt}$$ Hence I got: $$V=-Bvl$$ Now I found current: $$i=-\frac{Bvl}{R}$$ Putting $R=\frac{\rho l}{A}$, $\frac{i}{A}=j$, $\frac{1}{\rho}=\sigma$

I got $$j=-\sigma vb$$ Accounting for the directions, $$j=\sigma(-\vec{v}\times\vec{B})$$

So I got $$\vec{E}=-\vec{v}\times\vec{B}$$ But to my surprise, answer is given zero. Their explanation is that since railings are purely conducting, $\delta V=0$, but it seems very wrong to me, because we are dealing with rod only here, not the railings. So, please help here, and tell if I did some mistake on my part.

Answer 1

I think your problem comes from the writing of Ohm's law within a moving conductor: $$\vec{j} = \sigma(\vec{E}+\vec{v} \times \vec{B})$$ and not $\vec{j} = \sigma(\vec{E})$

Indeed, you must take into account the magnetic force that acts on the charge carriers.

On the other hand, I have the impression that there is a sign error in your calculation because you do not show your axes: you should find $\vec{j} =+ \sigma(\vec{v} \times \vec{B})$

You see that this leads to $ \vec{E}=\vec{0}$

In fact, the electric field must be zero because the rest of the circuit has zero resistance. There is no need for an electric field to move the charges there.

In a pictorial way, you can imagine the Laplace rail as a generator: within the rail, the magnetic force allows the charges to rise against the electric field. Charges accumulate at the ends of the rail and create an electric field.It is this electric field which allows the displacement of the charges in the rest of the circuit, which is purely ohmic.

If the circuit is open, the electric field cancels the magnetic component of the force and we find the Hall effect: $\vec{E}=-\vec{v} \times \vec{B}$

Hope it can help and sorry for my poor english.