How Do Faraday's Law and the EMF = BLv Equation Relate?

[mayer]

Alright, I need to put my question in the context of the probably ubiquitous example of a conductor rod moving perpendicularly over a magnetic field to generate an emf between the two sides of the rod. That emf is equal to BLv (magnetic field, length, velocity) according to my book(studying for MCAT). Now when that rod is placed within a closed circuit with a resistor retaining its ability to move back and/or forth to generate an electrical current, how does Faraday's Law relate to the emf = BLv equation. In other words, how do these two equations emf = delta(magnetic flux) / Area and emf = BLv relate? Are they essentially the same thing? Are they additive?

Thank You

 

[ShayanJ]

I recently posted something about this which I think gives you your answer. Take a look at here!

 

[milesyoung]

Say you have the rod moving on stationary conductive rails with the resistor fixed between their endpoints. According to Faraday's law of induction (its integral form), the emf induced in the circuit is given by:
$$
\varepsilon = -\frac{d\Phi}{dt}
$$
where $\Phi$ is the magnetic flux linking the circuit.

If the magnetic field is uniform and perpendicular to the plane of the circuit, you have:
$$
\Phi = B A(t)
$$
where $A(t)$ is the area enclosed by the circuit.

If $l$ is the length of the rod and $v$ is its velocity, the rate of change of $A(t)$ with respect to time is $l v$, so:
$$
\varepsilon = -\frac{d\Phi}{dt} = -\frac{d(B A(t))}{dt} = -B\frac{dA(t)}{dt} = -B l v
$$
The negative sign here just means that you have to assign the reference polarity of $\varepsilon$ opposite of what's shown in the derivation of your expression for the "motional emf" of a conductive rod.

In short, both approaches should give the same result. I'd guess your book makes use of the Lorentz force law in its derivation.

 

[mayer]

Ah yes thanks to the both of you. And yes these MCAT books summarize quite a bit and end up making it harder to understand.