# 15.2.2 Origin of Motional EMF

Faraday’s Law describes two distinct phenomena: Transformer emf and motional emf. Transformer emf ($\displaystyle \varepsilon =-\frac{{d\Phi }}{{dt}}=-NA\frac{{dB}}{{dt}}$) requires a magnetic field that changes with time (and a coil that does not have to move). Motional emf ($\displaystyle \varepsilon =-\frac{{d\Phi }}{{dt}}=-B\frac{{dA}}{{dt}}=-BLv$) requires a moving wire (and a magnetic field that can be constant).

Why does a changing magnetic field induce an emf in a coil? Well, there is not much to elaborate here without going into university physics. So it just happens, take it as. Why does moving a metal wire induce an emf? Ahh, this one does have a H2-level explanation: the emf arises from magnetic forces acting on moving charges. In other words, $\displaystyle \varepsilon =BLv$ happens because  $\displaystyle {{F}_{b}}=Bqv$!

A wire is neutral in charge as a whole of course. But a metal wire has a lot of mobile electrons in it! When the wire is moving, the electrons in it (plus the lattice ions) must also move with the wire. So when a wire moves across a magnetic field, each and every individual electron in the wire experiences the $\displaystyle {{F}_{b}}=Bqv$ force!

In the diagram above, a wire is moving rightward at speed v. So every electron in the wire experiences a downward magnetic force $\displaystyle {{F}_{b}}=Bqv$. Since these are free and mobile electrons, they will shift away from the top and towards the bottom end of the wire. This causes the top and bottom end to become positively and negatively charged respectively. As a result, a downward internal electric field E is formed in the wire.

Now that there is an internal E field, each electron in the wire experiences an upward electric force of $\displaystyle {{F}_{e}}=qE$ in addition to the downward magnetic force of $\displaystyle {{F}_{b}}=Bqv$. There is thus a self-stopping mechanism to the charge accumulation process. The excess electrons at the bottom will prevent more electron accumulation once

\displaystyle \begin{aligned}{{F}_{b}}&={{F}_{e}}\\Bqv&=qE\end{aligned}

Now since the electric field E is uniform (and $\displaystyle E=-\frac{{dV}}{{dx}}$), the electric potential gradient is simply $\displaystyle \frac{\varepsilon }{L}$, where e is the potential difference between the top and bottom ends of the wire, and L is the length of the wire. So

A metal wire of length L moving across a magnetic field B at speed v thus becomes a “magnetic battery” with emf of BLv. In a chemical battery, charges gain EPE through work done by “chemical” forces. In a moving wire, charges gain EPE through work done by magnetic forces. What’s cute about this “magnetic battery” is that it must keep moving in order to maintain the emf between its terminals.

Concept Test

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