I’m trying to understand the basic physical laws applying to Faraday’s induction experiment as pictured below. The Wikipedia site doesn’t go into detail.

When the smaller diameter inductor is inserted into the the larger, is the Lorentz Force what causes the electrons to move through the larger diameter inductor and through the voltmeter?

If so, I’m having trouble seeing how Lorentz applies. I point my middle finger in the direction which the inner inductor slides (magnetic field) B, my index finger seems like it should point also in the sliding direction v and now I’m confused because B and v should be perpendicular. I understand also that V = dA . dB/dt but curious if Lorentz applies as well.

Can anyone explain this experiment?

$$\epsilon = - \frac{d}{dt}\iint \vec{B} \cdot \vec{da}$$

The derivative is on the outside of the integral, this means an emf can be caused by a changing magnetic field, or a changing surface/moving wire.

Stationary wires:

When $$\vec{da}$$ is independant in time

$$\epsilon = \oint \vec{E} \cdot \vec{dl} = - \iint \frac{\partial \vec{B}}{\partial t} \cdot \vec{da}$$

This emf is directly due to the induced electric field.

In the absence of changing magnetic fields

When there is a moving loop

$$\epsilon = \oint \vec{v} × \vec{B} \cdot \vec{dl} = - \frac{d}{dt}\iint \vec{B} \cdot \vec{da}$$

In general:

When there is a changing surface, and a changing magnetic field

$$\epsilon = -\frac{d\phi_{B}}{dt} = \oint (\vec{E} + \vec{v} × \vec{B}) \cdot \vec{dl}$$

The proof of the decomposition of the emf in faradays law can be found here:

However I think a better way of looking at it, is by assuming the emf is caused by the full lorentz force, and then working backwards to Identify which part causes which (to derive the maxwell Faraday equation in the first place)

• B and V do not need to be perpendicular in general, why do you think this? Commented May 27, 2022 at 19:00

The Lorentz Force is the sum of the forces on a charge, $$q$$, due to the electric field, $$\mathbf E$$ and the magnetic field, $$\mathbf B$$ at the point in space where the charge is. Thus $$\mathbf F_{Lorentz}=q\mathbf E+ q\mathbf v \times \mathbf B$$ I call the first term on the right 'the electric Lorentz Force' and the second term 'the magnetic Lorentz Force'.

In the experiment of Faraday for which you give the picture, the large coil in which the voltage is shown to be induced is stationary, so the electrons in it have $$\mathbf v=0$$, and there is no magnetic Lorentz force on them. Therefore the induced emf is due to the electric Lorentz Force, $$q\mathbf E$$. The electric field, $$\mathbf E$$, is non-conservative and given by the Faraday-Maxwell equation (integrated form): $$\int _s \mathbf E.d\mathbf s=-\frac {\partial}{\partial t}\int_S\mathbf B . d\mathbf S$$ in which $$\mathbf {ds}$$ is an element of the line bounding the area of which $$\mathbf{dS}$$ is an element, and the line integral is taken all round the area.

We take a cross-section of the coil as the area, and note that $$\frac {\partial}{\partial t}\int_S\mathbf B . d\mathbf S$$ is non-zero because we are moving the small, current-carrying coil. So there is an electric field all round the (circular) boundary of this cross-section. This electric field urges electrons through the turns of the coil – the induced emf.

[Suppose that Faraday had kept the small current-carrying coil stationary, but had moved the large coil. In that case, for the free electrons in the large coil, $$\mathbf v \neq 0$$ and, moreover, $$\mathbf B$$ due to the small coil is not strictly parallel to the axis (the field lines 'splay out') so $$q\mathbf v \times \mathbf B \neq 0$$. But this time, $$\mathbf E = 0$$, because the $$\mathbf B$$ due to the small coil doesn't change with time.

It is not difficult to show that the emf is the same whether we calculate it using the electric Lorentz force in the case when the large coil is stationary and the small coil moving, or using the magnetic Lorentz force in the case when the large coil is moving and the small coil stationary. All that matters is the relative velocity – which is as it should be.]

• Should I be applying the cross product to vectors per some correspondence to the experiment diagram? Like I mention, it seems like the magnetic field and moving charge (In the wires of the tube moving) are the same vector resulting in 0 from the cross product. I think I’m viewing the geometry wrong
– Nick
Commented May 27, 2022 at 19:18
• (a) You should be applying the Lorentz force formula to the free electrons in the large coil. Are you doing so? (b) If you take any cross-section of the coil, there is a varying $\mathbf B$ (roughly parallel to the coil axis) through it due to the movement of the small, current-carrying coil. So there is an electric field all round the (circular) boundary of this cross-section. This electric field urges electrons through the turns of the coil – the induced emf. Commented May 27, 2022 at 19:30
• For OP: This experiment, from the lab frame, has nothing to do with the lorentz force. We are finding the EMF on a STATIONARY coil. The emf is due to the induced E field, as a result of the changing magnetic field from the moving coil. You shouldn't be touching the cross product in analysing this situation from the lab frame. Commented May 27, 2022 at 22:13
• "Like I mention, it seems like the magnetic field and moving charge (In the wires of the tube moving) are the same vector" the velocity of the charges in the moving loop are not in the same direction as the B field, you're probably.not considering the component of velocity as a result of the current. In any case, you shouldn't be analysing the lorentz force on the moving loop anyway. Commented May 27, 2022 at 22:20
• @Jensen Pauli "This experiment, from the lab frame, has nothing to do with the lorentz force. " I suggest that you check out what is meant by the Lorentz Force. It includes the force, $q \mathbf E$ experienced by a charge, even a stationary charge in an electric field. This is the force that drives the free electrons round a stationary coil through which the magnetic flux is changing. Commented May 28, 2022 at 8:04