About another thread, Faraday's law of induction

[fluidistic]

Homework Statement

Here's the situation: https://www.physicsforums.com/showthread.php?t=363089.
First, does a magnetic flux through a rod considered with no surface area makes sense? To reach the result, we have to assume the flux through it to be -BLR since it's going out from the magnetic source (namely an infinite wire).

Second question : Does that mean that when I push on the rod to make it closer to the wire, the emf induced in the rod acts to opposes the magnetic field due to the wire? In other words, does this induced emf creates a magnetic field that opposes the one of the wire, so that the total magnetic field decreases?
While if I pull on the rod outside from the wire, the emf is positive and hence it creates a magnetic field that will sum up with the one of the wire, making a greater magnetic field. Am I right on this? I've particularly the doubt that the induced emf will create a magnetic field, since there's no closed circuit.
Can someone explain to me what really happens in the situation of the problem?
Thank you very much, sincerely.

 

[jbsteele]

Homework Equations Faraday's Law: ε = -N*dφ/dt The Attempt at a Solution Yes, a magnetic flux through a rod with no surface area can make sense. The flux through it would be equal to -BLR since it is going out from the magnetic source (in this case, an infinite wire). As for your second question, it is correct that when you push on the rod to make it closer to the wire, the induced emf acts in such a way as to oppose the magnetic field due to the wire. This is because Faraday's Law states that ε = -N*dφ/dt, where ε is the induced emf, N is the number of turns, and dφ/dt is the change in flux over time. Since the flux is decreasing as the rod moves closer to the wire, the induced emf will be negative, thus creating a magnetic field that opposes the one of the wire and decreasing the total magnetic field. Conversely, when you pull the rod away from the wire, the induced emf is positive and thus creates a magnetic field that will sum up with the one of the wire, making a greater magnetic field.

 

[kerimek]

I would like to address the questions raised in the thread about Faraday's law of induction. First, it is important to understand that in order for Faraday's law to apply, there must be a change in magnetic flux through a closed loop or circuit. In the given situation, the rod does not have a surface area and therefore cannot have a closed loop or circuit. This means that the concept of magnetic flux through the rod does not apply.

Regarding the second question, it is correct that the induced emf in the rod will act to oppose the magnetic field due to the wire. This is known as Lenz's law, which states that the direction of the induced current will be such that it creates a magnetic field that opposes the change in magnetic flux that caused it. This is why, when the rod is pushed closer to the wire, the induced emf is negative, as it opposes the increasing magnetic field due to the wire. Similarly, when the rod is pulled away from the wire, the induced emf is positive, as it opposes the decreasing magnetic field.

It is also important to note that the induced emf does not create a magnetic field on its own. It is the current induced by the emf that creates the opposing magnetic field. This current can only exist if there is a closed loop or circuit, which, as mentioned earlier, is not present in this situation.

In summary, the concept of magnetic flux through the rod in this situation does not apply, and the induced emf and current will act to oppose the change in magnetic field caused by the wire. I hope this explanation helps to clarify the situation.