My mind is blown. ΔV problem from MIT lecture.

In summary: If the objects are connected with a wire then the field will push electrons into the wire and that will increase the PD. If the objects are connected with a resistor then the field will push electrons into the resistor and that will decrease the PD.
  • #1
mrspeedybob
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65
The lecture in question...


So what happens if I actually build this circuit and hook my fluke to it? How can moving my meter from the right side of the table to the left change my voltage reading if the circuit and my meter's connection to it is untouched? What happens if I put the meter directly on top of the solonoid?

Can someone try to explain this in a way that is different than the professor did?
 
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  • #2
Go read this thread:

http://forum.allaboutcircuits.com/showthread.php?t=16150&highlight=lewin
 
  • #3
mrspeedybob said:
The lecture in question...


So what happens if I actually build this circuit and hook my fluke to it? How can moving my meter from the right side of the table to the left change my voltage reading if the circuit and my meter's connection to it is untouched? What happens if I put the meter directly on top of the solonoid?

Can someone try to explain this in a way that is different than the professor did?


Good question.
I think the answer, in practical terms, must be to do with the way your meter leads would routed (there have to be connecting leads, of course). They will also experience an induced emf as the magnet moves in and out. If they are on the left of the loop, the emf will be in the opposite direction to the emf induced when they're on the right side - this will add or subtract the emf from the emf across the various components and give you different answers.
 
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  • #5
Jony130 said:

Great article. I love it when someone has time, inclination and resources to demonstrate such basic stuff. Lewin's problem is that he sounds a bit too much a a showman at times and that can lead to doubt in the minds of students.
Lesson: use Kirchoff with care, same as a chain saw.
 
  • #6
Thanks for the insights. This is making much more sense now.

Suppose though that we connect 2 metal objects to different points in the circuit. We pulse the solonoid. At the same instant, while the field is moving, we disconnect the 2 metal objects. Each one should now have a static charge representitive of the potential at the point it was connected to. We've taken the meter leads completely out of the picture. We measure the potential difference between the 2 objects after the circuit is powered down. What do we get?
 
  • #7
The PD between your two objects cannot be defined in that case without specifying the path you want to take. It's not a 'conservative field, as with a battery supply.
(Nice try! :wink:)
 
  • #8
Electrons in a wire do not behave like water. It's not the case that you push electrons into the wire and those then push along the other electrons.
All the electrons in the wire are pushed by an electric field and that field acts on the entire wire and all resistors in the circuit. That includes the wires that lead to the meter and the meter itself. It's all inside the electric field that is induced by the solenoid.
The same can be said about a battery connected to a circuit. The battery is not simply pushing electrons into the circuit at one end and pulling them out at the other. The battery produces an electric field that is felt by the entire circuit and can push electrons in the entire circuit simultaneously. Resistors will cause electrons to "pile up" so that they become a source of an electric field themselves. If you overlap the electric fields produced by all the resistors with the field produced by the battery and use that to calculate the potential differences inside the circuit you will get the same results you would get with the "normal" method i.e. kirchhoff as long as no changing magnetic fields are involved.
If you briefly connect two metal objects to point A and D while the magnetic field is changing the PD you get between them depends on how much charge is being pushed into them by the electric field. That in turn depends on how the objects are connected and in which direction the field is pointing.
 

1. What is the ΔV problem from the MIT lecture?

The ΔV problem is a concept discussed in a lecture at the Massachusetts Institute of Technology (MIT) that refers to the change in velocity required for a spacecraft to successfully complete its mission. This problem arises in space travel when a spacecraft needs to change its velocity to enter a different orbit or reach a different destination.

2. Why is the ΔV problem important in space travel?

The ΔV problem is important because it determines the amount of fuel and energy required for a spacecraft to complete its mission. The higher the ΔV, the more fuel and energy will be needed, making it a crucial factor in mission planning and spacecraft design.

3. How is the ΔV problem calculated?

The ΔV problem is calculated using the rocket equation, which takes into account the mass of the spacecraft, the mass of the fuel, and the specific impulse of the rocket engine. This calculation is essential for determining the amount of fuel needed for a spacecraft to reach its desired destination.

4. Can the ΔV problem be solved?

The ΔV problem can be solved by using various methods such as gravity assists, multi-stage rockets, and efficient propulsion systems. However, it is a complex problem and often requires trade-offs between fuel efficiency and mission objectives.

5. How does the ΔV problem affect space missions?

The ΔV problem has a significant impact on space missions as it determines the capabilities and limitations of a spacecraft. It influences the trajectory, duration, and success of a mission. Therefore, understanding and addressing the ΔV problem is crucial for the success of space exploration and travel.

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