Voltage generated in moving wire

hyperphysics:

When a conductor moves through a magnetic field, there will be a generated motional emf. This is one example of Faraday's Law and it arises from the magnetic force.

I don't understand how moving wire in a magnetic field is related to faraday's law. Every where i read about faraday's law, it talks about a change in magnetic flux through a closed surface (such as a loop wire). So how faraday's law can be applied to a single moving wire (open circuit) in a magnetic field?

So your problem is this one The wire has length $L$, it moves with constant speed $v$ in the $x$ direction and there is a constant magnetic field $\mathbf{B}$ to the right side of the red dotted line. First, ignore the green line I drew.

We know that every electron in the wire will experience a magnetic force

$$\mathbf{F}_m = q \cdot \mathbf{v}\times \mathbf{B} = -qvB \ \mathbf{\hat{y}}$$

where $q$ is the charge of an electron. This means that the negative charges (electrons) will move up, leaving behind a positively charge atom. So now there will be an accumulation of negative charge in the superior extreme of the wire, and an accumulation of positive charge in the inferior one. This creates an electric field that generates an electric force on those charges. So, in equilibrium, you have that

$$\mathbf{F}_m = \mathbf{F}_e \implies qE = qvB \implies fem = EL = vbL$$

So a voltage appears between the ends of the wire.

Now let's take into account the green curve I've drawn. This creates a closed path $C$, and the magnetic flux that goes through it will change in time as the bar moves. If you stand on the wire (i.e. now $v=0$, so that's your new reference system), if you solve Faraday-Lenz equation

$$fem = - \frac{d}{dt} \iint \limits_S \mathbf{B}\bullet d\mathbf{S}$$

where $S$ is the surface described by $C$ (namely, the wire together with the imaginary green curve), what do you get?

Hint: You should get exactly the same result as before.

• Suppose that wire is already inside magnetic field and it starts moving inside magnetic field. In this case there is no change in magnetic flux so Faraday's law tells that there is no emf but as wire moves, there is magnetic force on electrons in wire that forces them to move, i am right? – M3hRaN Jan 10 '17 at 20:13
• @M3hRaN You're right, I was mistaken. The green curve should be on the other side of the wire, and should always have the left border outside the magnetic field. – Tendero Jan 10 '17 at 20:20
• @Tendero I think Faraday's law talks about the flux through the body , the meat of the body. Not a hypothetical cuve – Shashaank Jan 13 '17 at 10:18
• @Shashaank Can't we use Maxwell-Faraday equation for such hypothetical rectangular ? I think Tendero meant Maxwell-Faraday equation, not just Faraday's law of induction. There is a similar Q/A about this, would you mind check whether the answer is correct ? l – M3hRaN Jan 13 '17 at 10:36
• @M3hRaN Answers can be correct. I have seen that post as well as the arguMents Tendero has given. In the link what if dx tends to 0. It should be the flux through the actual body. You yourself think then why Griffiths and Feynman don't present this argumeNt . – Shashaank Jan 13 '17 at 12:49

A short explanation would be that you shouldn't cite Faraday's law, but use the Lorentz-force to explain the effect. Since the charged electrons in the moving conductor have velocity perpendicular (or velocity with perpendicular component) to the field, the Lorentz-force will force them to move through the wire. (not exactly in that direction in the most cases, but normally they can not leave the conductor on the sides.) For the deeper meanings and causes you should read the material mentioned in Shashaank's answer, but this is a working surface explanation.

To detect the voltage induced in the wire, you must have a voltmeter connected to the two ends of the wire and thus forming a loop. You can show that the rate of change of the flux going through the loop is equal to Blv (if the variables are perpendicular).