Why does more rapidly changing magnetic field induce greater EMF?

I am confused about the above question. I know that when you move a magnet in and out of a close loop of wire, you induce a EMF in the wire. The change of moving the wire in a out of a wire is what is producing the wire. However, why is it that the faster you move the magnet in and out of the wire, the greater EMF you produce? I can't wrap my head on this because you have the same magnet pulling in and out with what I assume is a same magnitude of magnetic field. It's not as if the value of magnetic field is increasing. So why is it that the faster you pull the magnet in and out, the more EMF you induce? Thanks.

• The force on the electrons in the wire is proportional to their speed and to their to the strength of the magnetic field. This is how nature is. If the wire just sits in the magnetic field nothing happens. Another way of putting it is that the magnetic field interacts with a current not with a charge. – hyportnex Apr 16 '17 at 15:40
• Are you asking why it makes sense that Faraday's law is the way it is, or are you asking why Faraday's law leads to this conclusion? – user4552 Oct 3 '19 at 0:40

$\epsilon=-\frac{d\phi}{dt}$, where $\phi$ is the magnetic flux.

As you can see the the induced emf depends the the rate of change of magnetic flux. The faster you move the magnet, more will be rate of change of magnetic flux and thus, greater emf will be induced.

$$\epsilon=-\frac{d(\phi)}{d(t)}$$ $$\epsilon=BL\frac{dL_1}{dt}$$ $$\epsilon=BLV$$ This tells that that the increase of velocity is the increase of emf this equation was derived from the movement of a ramp inside a magnetic field (no negative sign because of cos (180)=-1).

why is it that the faster you move the magnet in and out of the wire, the greater EMF you produce? I can't wrap my head on this because you have the same magnet pulling in and out with what I assume is a same magnitude of magnetic field.

This is because the EMF in this case is due to induced electric field, not directly due to magnetic field. The faster the magnet moves, the higher the rate of change of magnetic field and the higher the induced electric field. This follows from the Faraday law of electromagnetism.

Consider the Faraday-Neumann-Lenz's Law: $$\mathcal{E}_{emf}=-k\frac{d\Phi_\Sigma(\vec{B})}{dt}$$ where $$\mathcal{E}_{emf}$$ is the electromotive force induced around the circuit $$\gamma=\partial\Sigma$$ (the circuit $$\gamma$$ is the border of a surface $$\Sigma$$) by the time rate change of the magnetic flux $$\Phi_{\Sigma}(\vec{B})$$ linking the circuit: $$\Phi_\Sigma(\vec{B})=\iint_\Sigma\langle\vec{B},\hat{n}\rangle d\sigma$$ Then: $$\frac{d}{dt}\iint_\Sigma\langle\vec{B},\hat{n}\rangle d\sigma=\iint_\Sigma\frac{d}{dt}\langle\vec{B},\hat{n}\rangle d\sigma=\iint_\Sigma \frac{d}{dt}[B(x(t),y(t),z(t))] d\sigma=\iint_\Sigma\left[\frac{\partial B}{\partial x}\frac{\partial x}{\partial t}+ \frac{\partial B}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial B}{\partial z}\frac{\partial z}{\partial t}\right]d\sigma=\iint_\Sigma\left[\frac{\partial B}{\partial x}\dot{x}+ \frac{\partial B}{\partial y}\dot{y}+\frac{\partial B}{\partial z}\dot{z}\right]d\sigma$$ Where in the first passage we exchanged the integral with the derivation since we took a fixed surface, in the second passage, we took a magnetic field which is always parallel to the surface's normal $$\Sigma$$ (you can think of a surface which is a plate and a uniform magnetic field perpendicular to the surface) and in the third equality we have used the chain rule.

It is clear that if the field $$\vec{B}$$ is time dependent implicitly through its coordinates, which in turn it means that we are moving our source of the magnetic field, then our electromotive force induced by this change in flux, depends on the velocity $$\dot{x},\dot{y},\dot{z}$$ of the magnetic field.

Consider two situations:

1. We have a loop of wire sitting at rest on a table, and we move the magnet towards it at a speed $$v$$.

2. We have a magnet sitting at rest on a table, and we move a loop of wire towards it at a speed $$v$$.

If you believe in the principle of relativity (and you should), you would expect the amount of current induced to be the same in both cases. Moreover, we know that the magnetic force on a charge is proportional to $$v$$. Thus, we would expect that in situation #2, the current induced in the loop would be larger when the speed of the loop is larger. But since we expect situations #1 and #2 to be the same, it must also be the case that a faster-moving magnet creates a greater current as well.

It should be noted that in intro-level electricity & magnetism classes, the descriptions of these two situations are different. In situation #1, the magnetic field at the location of the wire is changing, which means that there is an induced electric field, and this electric field is what gets the charges moving inside the wire. (The magnetic field can't do it, since the charges are initially at rest.) In situation #2, however, the magnetic field exerts a force on the moving charges in the wire, causing them to flow around the loop. Before Einstein came along, the fact that these two experiments gave exactly the same result by different mechanisms was basically viewed as a big coincidence; Einstein's famous paper describing special relativity cites this coincidence as evidence for his ideas.