Electric Force on a Charge in a Solenoid

AI Thread Summary
The discussion focuses on understanding the electric force on a charge within a solenoid, particularly in relation to the induced electric field due to a changing magnetic flux. Participants clarify that the direction of the induced EMF is crucial, as it relates to the right-hand rule and the concept of Faraday's law of electromagnetic induction. The key point is that the electric field experienced by the charge is induced by the changing magnetic field, and when the charge is stationary, the velocity term becomes zero. The conversation emphasizes the importance of recognizing the time-dependent nature of the magnetic field in determining the direction of the force on the charge. Overall, the interaction between the electric field and the magnetic field is central to understanding the problem.
Helly123
Messages
581
Reaction score
20

Homework Statement


http://[url=https://ibb.co/dgUy6T]https://preview.ibb.co/iyqS0o/20180525_213806.jpg
[ATTACH=full]226151[/ATTACH]
Since i only know the field direction, increasing go into page. Why the answer is C?

[ATTACH=full]226152[/ATTACH]
Why the answer "a" ?

The R and r on the pic is respected to what?

[h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2]
Right hand rule : finger tip = current
Thumbs = force
Palm = B

Or
Finger tip = velocity
Thumbs = current
Palm = B
 

Attachments

  • 20180525_213806.jpg
    20180525_213806.jpg
    26.5 KB · Views: 1,047
  • 20180525_213745.jpg
    20180525_213745.jpg
    27.4 KB · Views: 770
Physics news on Phys.org
The key word here in the statement of the problem is "increasing" and they should say "increasing with time". The question involves the Faraday EMF that occurs. The changing magnetic flux creates an EMF. They basically are asking the direction of the induced ## E ## of the EMF.
 
  • Like
Likes Helly123
Could it be like this?
20180526_153723.jpg


20180526_153612.jpg


Force direction the same as Emf's ?
 

Attachments

  • 20180526_153723.jpg
    20180526_153723.jpg
    29.3 KB · Views: 747
  • 20180526_153612.jpg
    20180526_153612.jpg
    31.4 KB · Views: 686
That is correct. The EMF is the integral of an induced electric field along the path. The electric field ## \vec{E}=\vec{E}_{induced} ## is what the positive charge responds to, and experiences a force ## \vec{F}=q(\vec{E}+\vec{v} \times \vec{B}) ##. ## \\ ## (And of course, when they say "placed at point b", etc. that means ## \vec{v}=0 ## so that ## \vec{v} \times \vec{B}=0 ## ).
 
  • Like
Likes Helly123
Charles Link said:
That is correct. The EMF is the integral of an induced electric field along the path. The electric field ## \vec{E}=\vec{E}_{induced} ## is what the positive charge responds to, and experiences a force ## \vec{F}=q(\vec{E}+\vec{v} \times \vec{B}) ##. ## \\ ## (And of course, when they say "placed at point b", etc. that means ## \vec{v}=0 ## so that ## \vec{v} \times \vec{B}=0 ## ).
Thank you @Charles Link
 
  • Like
Likes Charles Link
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
Back
Top