Faraday's Law with Acceleration

AI Thread Summary
The discussion revolves around the application of Faraday's Law in scenarios involving acceleration of a circuit moving through a magnetic field. It highlights the confusion regarding the change in magnetic flux when the circuit accelerates, despite the magnetic field being constant. Participants clarify that even without current flow, accelerating electrons within the circuit can induce an electric field, leading to a time-varying magnetic field. The conversation emphasizes the mathematical relationships involved, particularly the role of velocity and acceleration in determining induced electromotive force (EMF). Ultimately, the participants seek to understand how these principles apply to their specific scenario of a circuit in motion.
KabooHahahein
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I decided to join this forum because we were baffled on the following fact made by a professor. We were hoping to get this cleared up here, since the prof said the math was complicated.

Let's assume any arbitrary circuit (forming an area) moving at velocity v through a perpendicular field B to the plane of the area of the circuit. Now, the change in magnetic flux is 0 in constant v. However, it was pointed out that it is not 0 in any acceleration.

Our prof pointed out that it can be imagined as an increasing area as a magnetic lawnmower increases in speed.

How are the mathematics applied to this sort of problem?

Thank you!
 
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This is not the section to ask help for homeworks or assignments but I'll give you a go for this time. Please use the dedicated section for these kinds of purposes from now on.

You need to be more specific ;

Is the circuit is completed ?
Are there any currents flowing in the circuit ?
If so what are the specs ?

It has nothing to do with the area of the pcb or the circuitary. Your professor probably asked for this ;

Even if there is no current flow in the circuit, due to electrons in the pcb traces and the wiring, when the whole system accelerates you will be accelerating the electrons with the system. A moving charge will induce an electric field and a time varying electric field will induce a magnetic field . Taking derivative of velocity will give you acceleration and taking curl of electric field will give you minus derivative of magnetic field. From now on it's convenient to express ;

F = q.v x B

q is the coulomb of your charge, v is velocity and B is magnetic flux density vector.

Although this might not be the case at all, depending on what you really mean.
 
The circuit is completed, and we're talking about the induced EMF (resulting in an induced current) from Faraday's Law from a change in flux. Initially, I would assume that there are no currents on the circuit.

Now our issue is that the flux is (B)(A), and that it is constant under an externally generated constant magnetic field. Shouldn't the change in flux be 0, no matter whether the circuit is at constant velocity or acceleration through this constant field?

I apologize for making this look like homework. This is not anything assigned, we were simply curious as how it's possible that under a changing velocity, the change in magnetic flux through the loop would not be zero.
 
KabooHahahein said:
The circuit is completed, and we're talking about the induced EMF (resulting in an induced current) from Faraday's Law from a change in flux. Initially, I would assume that there are no currents on the circuit.

Now our issue is that the flux is (B)(A), and that it is constant under an externally generated constant magnetic field. Shouldn't the change in flux be 0, no matter whether the circuit is at constant velocity or acceleration through this constant field?

I apologize for making this look like homework. This is not anything assigned, we were simply curious as how it's possible that under a changing velocity, the change in magnetic flux through the loop would not be zero.

If flux (B) is constant you have a time-independent electric field also implying there is DC operation going on in the circuit.

There is no such thing as (B)(A).

\int\int (dB / dt) ds = \oint (\nablax E) ds
 
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