Quantcast

A community for students. Sign up today!

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

modphysnoob

  • one year ago

EMF

  • This Question is Closed
  1. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1362247626272:dw|

  2. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so we are supposed to find current in the loop assuming theere is a resistor there so I will start off with using magnetic force forumla q v X B but the book set it equal to Electric force

  3. prakharluv
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    is there are any dimensions of loop?

  4. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]

  5. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1362247929761:dw|

  6. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    But \[EMF = BLv = IR \] so the current \[ I = \frac{BLv}{R} \]

  7. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I see, so there is no point to be messing around the force at all?

  8. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    No, not really. What book is this?

  9. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Fundamental of Applied Electromagnetics by ulaby

  10. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I am unfamiliar with it, sorry. I don't know, unless there's more to the question.

  11. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I will attach that page

  12. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

  13. prakharluv
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok for this loop,,, The magnetic flux linked with this loop will be, \[\phi=Blx\] Since x is changing with time, the rate of change of flux \[\phi\] will be induce emf given by : \[e=\frac{ -d \phi }{dt }= \frac{ d }{ dt } Blx\] \[e= -Bl \frac{ dx }{ dt }\] e=Blv

  14. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh, I see. What they are saying is the magnetic force on the particles is the same as if there was an electric field \[\vec{E} = (\vec{v}\times \vec{B}) \]

  15. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how did they come to that assumption?

  16. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so magnetic field here would push all positive particle down , negative particle up , between them is potential difference

  17. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yes. And they didn't come to any assumptions, it's fairly clear. There is a force acting on them: \[\vec{F} = q\vec{v}\times \vec{B} \] we know that for electric fields, \[\vec{F} = q\vec{E} \] so our situation is just like what we would find if there happened to be an electric field present \[\vec{E} = \vec{v}\times \vec{B} \]

  18. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I understand the formula , the part that confused me is why these two force must be equal

  19. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh, I think I understand , it is one force , not two force; we are just looking at it as if it two way

  20. Jemurray3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    mhmm

  21. modphysnoob
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    not quite?

  22. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Ask a Question
Find more explanations on OpenStudy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.