A student in my physics class posted, in a group, a wrong answer to a question.

The situation was: A plane has a wire extended between the tips of its wings and flies through a magnetic field, perpendicularly, while accelerating.

The question was: What will be the induced current?

His answer was 0, and he explained that as the plane was flying through a uniform field, there would be no change in Magnetic flux density, so no change in flux linkage and ultimately no induced current.

I stated, after clarifying with a question on here earlier today, that instead of consulting Faraday's laws, he should think about Fleming's left hand rule.

My teacher told me I was wrong, and said as the plane is accelerating, it cuts the field lines at a greater rate, so according to Faraday's law, a voltage will be induced, but he also says this voltage will be increasing.... Even if it was Faraday's laws, as this acceleration was constant, wouldn't the voltage be constant?

My next response is a lengthy one:

Faraday's law dictates proportionality between an induced EMF and rate of change of flux linkage. The rule of a wire cutting through fields lines is contrarily a result of Fleming's left hand rule, where a magnetic force is exerted on the electrons inside the wire that is cutting the field, in the direction of the wire; this leads to a potential difference between both sides of a wire, or induced current for a closed circuit. Also, flux linkage, and thus Faraday's law, refer to solenoids, don't they? Ultimately, the induced voltage will indeed increase due to the increase in the rate at which the wire cuts the field, but this isn't to due with Faraday's law which relates the change in flux linkage on a 2-dimensional conductor. the students understandable reasoning for his answer of an induced EMF of 0 came from him observing that the flux linkage is of constant magnitude, so will have 0 rate of change, no matter the acceleration. Using the left hand rule, and the fact that magnetic force, F, is the sum of the BQv, it can be seen that the electrons will be moved by a greater force, making for a greater potential difference, EMF or current, as the wire accelerates.

I also realised that with constant acceleration the voltage would be constant, if this was to do with Faraday's law, and will add that to what I have say.

I'm asking as I want to help this person, while not feeding them false information, wrongfully undermining my teacher and ultimately embarrassing myself.


There are two questions here: is there an EMF in the wire and is there a current in the wire? You calculated the force on the charges in the wire as $$F = qvB.$$ The EMF across the wire can be calculated from this: $$\varepsilon = \int_0^L E\,dx = \frac{1}{q}\int_0^L F\,dx = vBL$$ where $L$ is the length of the wire/wing. So, there is a non-zero EMF. But, because the wire is not a closed circuit, there is nowhere for the charges to go once they hit the end of the wire. If the plane is traveling at a constant speed, then the electrons in the wire will shift a bit until the total EMF, both from the magnetic field and the charge distribution, is zero. Then, the current will stop.

As the plane accelerates, the charges will continue to shift to one side, creating a current, but the current will not be constant. Judging by some initial scribbles of mine, the expression for the current along the wire will be a complicated function of position along the wire and time.


This is a case of motional EMF in uniform magnetic field with no induced electric field. The flux linkage concept is superfluous and irrelevant - there is no solenoid or similar coiled path of relevance and magnetic flux through any closed curve of fixed geometry is constant. Unfortunately many textbooks and teachers think that the Faraday formulation with magnetic flux and EMF for closed paths is general and use it also in scenarios where it is quite confusing and artificial, such as for motional EMF in open circuit. This muddles things and causes problems for student's understanding as motional EMF is very different from induced EMF.

Motional EMF is present in conductor when conductor moves in direction not aligned with magnetic field vector; this EMF is present even if magnetic flux for any fixed size loop is constant and no induced electric field is present. This EMF is due to internal forces of the rest of the conductor pushing on mobile charges inside.

If the conductor moves non-relativistically, motional EMF for a path $\gamma$ that goes from one end of the wire A to the other end B through the conductor can be approximately calculated as

$$ EMF = \int_\gamma \mathbf v \times \mathbf B \cdot d\mathbf s. $$

where $\mathbf v$ is velocity of the conductor (not velocity of mobile charges in it).

The forces behind this EMF cause quick redistribution of electric charge on the conductor and while that is happening, increasing electrostatic field and voltage between the ends. The process quickly stops when the force effect of EMF is counteracted by the established electrostatic field due to separated charges. In that equilibrium, voltage magnitude is the same as the EMF:

$$ |V_2 - V_1| = |EMF|. $$

In the simplest case where the wire is a straight piece of length $L$, moving perpendicularly to magnetic field, the integral can be simplified and we get

$$ |V_2 - V_1| = |EMF| = vBL. $$

If velocity is increasing in time slowly enough, then the motional EMF and voltage magnitudes are still given by this formula, just $v$ is a function of time. There will be a small electric current in the wire if the plane is accelerating or decelerating, but it is very weak and usually can be neglected, because very little current is needed to establish and modify the described charge separation.


I want to provide a more Faraday's Law point of view (in contrast to Lorentz force law PoV).

First, the description of Faraday's Law as a wire cutting through magnetic field is not great. Because such conception has many exceptions. A famous exception is: consider a magnet rotating and a piece of metal disk rotating in front of it. Lorentz Force law predicts a EMF while Faraday's Law doesn't.

I digressed, as to your question about the constant acceleration vs constant magnitude of EMF. I don't think that's accurate. Faraday's Law gives that the EMF is proportional to the rate at which magnetic flux is swept across by the wire. So it should be linearly related to velocity instead of acceleration.

The student you mentioned is not separating induced EMF with motional EMF. In the case of the plane, because there's no magnetic flux change, induced EMF is zero. However, motional EMF is not zero.

Induced EMF is given by Maxwell-Faraday's Law (different from Faraday's Law!), motional EMF is given by Lorentz Force law. However, Faraday's Law (at least a version of it) combines the two and gives the net EMF.

For more, you can see this post I wrote: https://physicintuited.wordpress.com/2021/05/07/why-faradays-law-is-weird/

Hope it helps!


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