Electric Force on a Charge in a Solenoid

  • #1
Helly123
581
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: 1,020
  • 20180525_213745.jpg
    20180525_213745.jpg
    27.4 KB · Views: 732
Physics news on Phys.org
  • #2
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
  • #3
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: 709
  • 20180526_153612.jpg
    20180526_153612.jpg
    31.4 KB · Views: 642
  • #4
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
  • #5
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
Back
Top