What exactly is happening to the electrons inside the wire while it is moving quickly downwards within the magnetic field?

Why do electrons move in one direction if the wire is being pushed down?

I don't care what direction electrons move. I want to know why they move in one direction?

enter image description here

  • 2
    $\begingroup$ If you read this you could ask more specific questions en.wikipedia.org/wiki/Electromagnetic_induction $\endgroup$
    – user108787
    Sep 28, 2016 at 14:20
  • $\begingroup$ @CountTo10 I didn't get the answer there. $\endgroup$
    – m0Onfang
    Sep 28, 2016 at 14:57
  • $\begingroup$ Maybe wire is not being push down but electrons feel the lorentz force? $\endgroup$
    – Ice-Nine
    Sep 28, 2016 at 16:15
  • $\begingroup$ The Perhaps it is helpful for you to read Why the electrons intrinsic spin and the related magnetic dipol moment are responsible for induction processes. The main point is that "This is possible only because the magnetic dipole moment and the intrinsic spin of the electrons have all the same dependence. (Protons for example are anti-aligned in relation to electrons.)" $\endgroup$ Sep 28, 2016 at 19:32

4 Answers 4


The primary reason for the movement of electrons is because of the Lorentz force given by F = q.v x B Now, this force is primarily acting because of the velocity of the electron and an external magnetic field acting on it. You may think why is magnetic field applying a force on an electrical charge? The reason is simple and elegant (took many years to realise and to further mathematically express and document the relations). Force in a static frame of reference (relative to the electron) will be solely due to an electric field, but when the electron is moving with respect to our frame, the electric field has to translate into a magnetic field, which is evident in the Lorentz force equation. Thus, an electric field due to the electron at rest will translate into a magnetic field in motion. But this is due to the electron, and of course, the electron will not experience a force because of the fields it itself produces, but this does give you an intuition that what an electric field is in a static situation is same as a magnetic field in a dynamic situation, thus both applying the same force. So, in your question, the field is from the left to right, while the velocity is downwards. Thus according to the cross product (keeping in mind the negative charge of the electron), a force will act in the forward direction on the electrons. Therefore, the front facing side of the wire will be having a positive potential with respect to the farther side.


Usually, Faraday's law of induction is derived from Maxwell-Faraday law. This law states that

$$ \nabla \times \mathbf{E} = -\frac{\textrm{d}}{\textrm{d}t}\mathbf{B}$$

Now, let's integrate this on the surface $S$ delimited by the closed loop $\Gamma$. We define the electromotive force $\mathcal{E}$ to be the circulation of $\mathbf{E}$ along $\Gamma$, so using Stokes theorem we get

$$ \iint\limits_S (\nabla\times\mathbf{E}) \cdot\textrm{d}\mathbf{S} = \oint\limits_\Gamma \mathbf{E}\cdot\textrm{d}\mathbf{l} = \mathcal{E} $$

and Maxwell-Faraday law becomes

$$\mathcal{E} = -\iint\limits_S \frac{\textrm{d}\mathbf{B}}{\textrm{d}t} \cdot \textrm{d}\mathbf{S}$$

Now, we would like to get the derivative out of the integral, however we can only do that (mathematically speaking) if the surface $S$ is constant. So, in the case of a fixed circuit, with $\Phi$ the flux of $\mathcal{B}$ through $S$,

$$\mathcal{E} = -\frac{\textrm{d}}{\textrm{d}t}\iint\limits_S\mathbf{B}\cdot\textrm{d}\mathbf{S} = -\frac{\textrm{d}\Phi}{\textrm{d}t} $$

which is Faraday's law of induction, in the case of a fixed circuit. If $S$ is not constant, then we can't switch $\iint\limits_S$ and $\frac{\textrm{d}}{\textrm{d}t}$, so we can't use Maxwell's law.

First of all, we need to understand what the electromotive force $\mathcal{E}$ is. It is defined as the circulation of $\mathbf{E}$ along $\Gamma$, but why ? In fact, we should instead define it a the work done - in the frame of reference of the conductor - per unit of charge along the circuit. Electrons in the conductor undergo the force

$$ \mathbf{F} = q(\mathbf{E}_{conductor} + \mathbf{v}_e\times\mathbf{B}_{conductor})$$

so the work done on one of them is

$$ W = \oint\limits_\Gamma \mathbf{F}\cdot\textrm{d}\mathbf{l} = q\oint\limits_\Gamma \mathbf{E}_{conductor} \cdot \textrm{d}\mathbf{l} + q\oint\limits_\Gamma (\mathbf{v}_e \times \mathbf{B}_{conductor}) \cdot \textrm{d}\mathbf{l} $$

However, the global motion of electrons in a wire is in the direction of the wire, so $\mathbf{v}_e$ and $\textrm{d}\mathbf{l}$ are colinear, which gives $(\mathbf{v}_e \times \mathbf{B}_{conductor}) \cdot \textrm{d}\mathbf{l} = 0$. Finally,

$$ W = q\oint\limits_\Gamma \mathbf{E}_{conductor} \cdot \textrm{d}\mathbf{l} $$

and we get back to the previous definition of $\mathcal{E}$.

Secondly, we will find the relation between the electromagnetic field expressed in two different frame of reference. Let's consider two frame $\mathcal{R}_1$ and $\mathcal{R}_2$, with $\mathcal{R}_2$ translating with a speed $\mathbf{V}$ relative to $\mathcal{R}_1$. Let's consider a charge $q$ with a speed $\mathbf{v}_1$ (resp. $\mathbf{v}_2$) in $\mathcal{R}_1$ (resp. $\mathcal{R}_2$). Since the electromagnetic force is a real force, its value doesn't depend on any frame of reference, so

$$\mathbf{F} = q(\mathbf{E}_1 + \mathbf{v}_1 \times \mathbf{B}_1) = q(\mathbf{E}_2 + \mathbf{v}_2 \times \mathbf{B}_2) $$

Also, we know that if we don't take the effects of special relativity into account $$\mathbf{v}_2 = \mathbf{v}_1 + \mathbf{V}$$

which gives finally

$$ \mathbf{E}_1 + \mathbf{v}_1 \times \mathbf{B}_1 = \mathbf{E}_2 + \mathbf{v}_1 \times \mathbf{B}_2 + \mathbf{V} \times \mathbf{B}_2 $$

or, equivalently,

$$ \mathbf{E}_1 - (\mathbf{E}_2 + \mathbf{V} \times \mathbf{B}_2) = \mathbf{v}_1 \times (\mathbf{B}_2 - \mathbf{B}_1)$$

This equation is true for any value of $\mathbf{v}_1$, so in particular we can take $\mathbf{v}_1$ and $\mathbf{B}_2 - \mathbf{B}_2$ colinear, so we finally get

$$\begin{cases} \mathbf{B}_2 = \mathbf{B}_1 \\ \mathbf{E}_2 = \mathbf{E}_1 - \mathbf{V}\times\mathbf{B}_1 \end{cases}$$

Notice that this result is only valid for $V \ll c$, but it won't be an issue in our case...

Now, we can find what the law of induction is in the case of a moving circuit placed in a constant magnetic field. Let's consider a deformable closed wire $\Gamma$ where each portion $[l,l+\textrm{d}l]$ has a speed $\mathbf{V}(l)$, placed in a constant magnetic field $\mathbf{B}$, then ,using the definition of $\mathcal{E}$,

$$\mathcal{E} = \oint\limits_\Gamma \mathbf{E}_{wire}(l) \cdot \textrm{d}\mathbf{l}$$

however, thanks to the previous paragraph,

$$\mathbf{E}_{wire}(l) = 0 - \mathbf{V}(l) \times \mathbf{B}(l)$$

so $$ \mathcal{E} = - \oint\limits (\mathbf{V}(l) \times \mathbf{B}(l)) \cdot \textrm{d} \mathbf{l} $$

Let's write $\mathbf{V}_N$ (resp. $\mathbf{V}_T$) the normal (resp. tangential) part of $\mathbf{V}$, then $\mathbf{V}_T$ and $\textrm{d}\mathbf{l}$ are colinear, so

$$(\mathbf{V}\times\mathbf{B})\cdot \textrm{d}\mathbf{l} = (\mathbf{V}_N\times\mathbf{B})\cdot \textrm{d}\mathbf{l} $$

Let's write $\mathbf{n}$ (resp. $\mathbf{t}$) the direction of $\mathbf{V}_N$ (resp. $\mathbf{\textrm{d}\mathbf{l}}$) and let $\mathbf{p}$ be so that $(\mathbf{t},\mathbf{n},\mathbf{p})$ is a direct orthonormal base. Now,

$$\begin{align} (\mathbf{V}_N\times\mathbf{B})\cdot \textrm{d}\mathbf{l} &= (\mathbf{V}_N \times (\mathbf{B}_T + \mathbf{B}_P))\cdot \textrm{d}l\, \mathbf{t} \\ &= (-V_N B_T \mathbf{p} + V_N B_P \mathbf{t} ) \cdot \textrm{d} l\, \mathbf{t} \\ & = V_N B_P \textrm{d}l \ \end{align}$$

However, $V_n\textrm{d}l\textrm{d}t$ is the area swept by the wire during $\mathbf{d}t$, and thus $V_n\textrm{d}l\textrm{d}t \times B_P$ is the magnetic flux swept during $\textrm{d}t$, so

$$\oint\limits_\Gamma V_N B_P \textrm{d}l = \frac{D\Phi}{Dt}$$

where I use $\frac{D\Phi}{Dt}$ to denote the flux swept per unit time, and, finally, we get the law of induction:

$$\mathcal{E} = - \frac{D\Phi}{Dt} $$

In the case of a simple wire falling, $\frac{\textrm{d}\Phi}{\textrm{d}t} = \frac{D\Phi}{Dt}$, however that's not always true (consider for example a rotating wire...).


When the magnetic flux through the wire changes, as it moves down, Faraday's law of induction says that it acquires an electromotive force.

The induced electromotive force in any closed circuit is equal to the rate of change of the magnetic flux enclosed by the circuit:

$${{\mathcal {E}}=-{{\mathrm d\Phi _{\mathbf {B} }} \over\mathrm dt}\ }$$

where ${ {\mathcal {E}}}$ is the $\rm EMF$ and $\Phi_{\mathbf B}$ is the magnetic flux.

Why do electrons move in one direction if the wire is being push down? I don't care what direction electrons move... I want to know WHY they move in one direction?

enter image description here

The EMF will now cause a current to flow in the external resistor R. This means that a similar current (electron flow) flows through the rod PQ itself, giving a magnetic force to the left.

If you read the above, and remembering that the direction of the electromotive force is given by Lenz's law,  which states that an induced current (the electrons) will flow in the direction that will oppose the change which produced it.


The flow of current is nothing but movement of electrons when a potential difference is applied across the ends of a conductor(in your case a wire). Electrons flow towards positive terminal and create a magnetic field around a wire. This magnetic field will cancel any other magnetic field that is of lesser or equal magnitude, which is why electrons being unaffected will keep moving along in the same direction.


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