In part (b) of this question, they have stated that the current that flows through the aeroplane is 0.00 A. When you use the value for emf from part (a) and the resistance, you can calculate that I = V/R = 0.490/1.78 = 0.275 A. I don't understand how they concluded that there must be no current in this situation; they must have also concluded it without doing any calculations for the question was only one mark. enter image description here enter image description here

Passing through the earth's magnetic field This link says that a current is indeed produced, which doesn't answer my question.
What is the effect of Earth’s magnetic field on airplanes? Here, they have stated that the resistance is great for a current. However, as shown above, the current in this question is still big enough for it to have an answer to 3 signficant figures, so I believe that this question is not assuming that the resistance is too large.
Emf induced on the wings of an airplane This article seems the most relevant. It says that the magnetic flux is not changing so there are no eddy currents created. So does that mean that the emf caused by the movement of a conductor through a magnetic field (ε=lvB) never produces current? It has to be emf caused by Faraday's law which causes a current?

Also, can you explain it through the fact that if the magnetic field is producing a force pushing the electrons to one side of the plane, the electrons can't travel the other way to go round the circuit and produce a current?

  • $\begingroup$ The wire that you use to connect the wings is also in the magnetic field. What's the EMF on the wire? $\endgroup$
    – BowlOfRed
    Commented Sep 21, 2023 at 5:44
  • $\begingroup$ The same as the plane? $\endgroup$
    – cabbagesss
    Commented Sep 21, 2023 at 6:03
  • $\begingroup$ So the net emf is? $\endgroup$
    – Farcher
    Commented Sep 21, 2023 at 7:22
  • $\begingroup$ 0.490V still but it didn't make sense to me why there could be an emf but no current $\endgroup$
    – cabbagesss
    Commented Sep 21, 2023 at 8:30
  • 1
    $\begingroup$ That is not the net emf in the complete circuit. You have a loop travelling through a region of constant magnetic field. $\endgroup$
    – Farcher
    Commented Sep 21, 2023 at 11:17

1 Answer 1


The force on an electric charge $q$ is given by the Lorentz force equation

$\vec F=q \vec E + q \vec v \times \vec B$

where $\vec E$ is an electric field, $\vec B$ the magnetic field and $\vec v$ the velocity of motion of the charge. The movement of the free electrons in the metal wing through the magnetic field leads to a force that displaces the electrons towards one end of the wing, which becomes negatively charged, and leaves a net positive charge at the other end. The movement of electrons represents a current flow in the wing but this occurs only for a very short time (when the plane first starts moving or when it first flies into the magnetic field). Very quickly the charge difference builds creating an electric field $\vec E=-\vec v \times \vec B$ that eventually balances the magnetic force, whereby $\vec F=0$. This opposing electric field results in a zero net current flow in the wing but creates a potential difference. It is this potential difference, also known as an electromotive force $\mathcal E=\int \vec E \cdot d \vec l$, that is calculated in the problem.

With the wire in place, nothing changes because the electrons in the wire experience exactly the same magnetic force and create exactly the same electric field that cancels it. So the total force on the electrons is zero and no current will flow.

However, you can get a current flow around a loop of wire in a magnetic field if the field strength passing through the loop is changing with time (strictly it is the changing magnetic flux that induces the EMF). Thus if the aeroplane was moving through a magnetic field gradient you could get a current flowing in the loop. In effect, the magnetic force on the wire would be different from that on the wing on account of the difference in the magnetic field strengths.


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