# 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. Apr 16, 2017 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, 2019 at 0:40

According to Faraday's law,

$\epsilon=-\frac{d\phi}{dt}$, 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.

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.

The induced EMF is a result of charge separation inside the rod which is a result of the free electrons being moved (by the BQv or the Lorentz's force) towards one end of the rod. Hence the faster the rod (and hence the electrons) is moved through the magnetic field, the greater the BQv forces on the electrons and greater the charge separation giving rise to a greater EMF. Hope this helps 🙂

$$\epsilon=-\frac{d(\phi)}{d(t)}$$ $$\epsilon=BL\frac{dL_1}{dt}$$ $$\epsilon=BLV$$ 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).

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.

I think you will find this most interesting, For example, let's say that I have a charged ring of charge density $$\lambda$$and in this ring there is a uniform magnetic field.

where the B field is perpendicular to the R vector of the ring path

when there is a changing magnetic field through this ring, $$\partial B/\partial t$$, faradays law states that there is an EMF in the ring.

Due to symmetry the electric field points parrallel to the ring at all times. Thus, when there's a changing magnetic field , the charge on the ring experiences a force, causing it to rotate ( as the ring itself is charged and are bound together)

Clearly faradays law states that if there's a magnetic field that starts from B0 and goes to 0 in 1 second, then there is a higher EMF in the ring than if it took 2 seconds to go to zero. So it is clear that each dq of charge $$\lambda dl$$ experiences a higher force, and thus the ring experiences a higher torque than if it took longer for the magnetic field to go to 0,

However! because it took less time to go to zero, it experiences a larger force FOR A SHORTER TIME. thus, the total amount of angular momentum imparted to the ring is the SAME In both cases and is independent of how fast the B field turns off an on, its just in one case the torque is higher so it rotates to some fixed$$\omega$$ faster

Proof?

for a circular ring, of radius a ,in the presence of a uniform magnetic field B that just encloses the ring the flux is $$\phi= B\pi a^2$$ $$\partial\phi/\partial t = \partial B/ \partial t \pi a^2$$

As per faradays law this is$$-\int E \cdot dl$$ $$-\partial B/ \partial t \pi a^2 = \int E \cdot dl$$

But what is the magnitude of the torque? General formula is $$\int R×E\lambda dl$$

For a circle, |R×E| is RE, Likewise R is just a, and are both independent of the integral so torque(N) becomes

N = $$\lambda a \int E dl$$

For symmetries of the problem$$E \cdot dl$$ is E dl, thus substituting E.dl into our torque equation gives

$$N = \lambda a * ( -\partial B/ \partial t \pi a^2)$$

$$N= -\lambda \pi a^3 \partial B/\partial t$$

N $$\partial t = - \lambda \pi a^3 \partial B$$

$$\int N \partial t$$= angular momentum

$$\int_{0}^{T1}N\partial t = \int_{B0}^{0} -\pi \lambda a^3 \partial B$$

Angular momentum = $$B0 \pi \lambda a^3$$

notice the formula for angular momentum is independent of $$\partial B /\partial t$$and only depends on the starting value of B

Thus no matter how fast you go from B0 -> 0 the same amount of angular momentum is imparted in both cases , its just one takes longer for it to happen. ( or e.g with an ammeter there is a larger current for a SHORTER time) Thus everything is conserved even though it may look like one has more momentum/ energy

The Simple Answer to this question, is that when we move the magnet , we are changing the strength of the Magnetic Filed.

This then affects the Magnetic Flux, which is the rate of change of the Magnetic Field. Which has a Formula of :

Magnetic Flux = Strength of magnetic Field x Area of the coil x cosine(theta)

Where theta is the angle between the Magnetic Field, and the line perpendicular to the face of the coil.

Now because the Magnetic Flux is lesser, the EMF produced would be lesser too

• EMF produced depends on rate of change in magnetic flux and not the value of the magnetic flux. Also, magnetic flux is the surface integral of magnetic field over an area, not the rate of change of the magnetic field. Sep 15, 2020 at 17:01