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.
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Homework Statement


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

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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
 

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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.
 
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Could it be like this?
20180526_153723.jpg


20180526_153612.jpg


Force direction the same as Emf's ?
 

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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 ## ).
 
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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
 
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