## modphysnoob 2 years ago EMF

1. modphysnoob

|dw:1362247626272:dw|

2. modphysnoob

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

is there are any dimensions of loop?

4. Jemurray3

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

|dw:1362247929761:dw|

6. Jemurray3

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

7. modphysnoob

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

8. Jemurray3

No, not really. What book is this?

9. modphysnoob

Fundamental of Applied Electromagnetics by ulaby

10. Jemurray3

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

11. modphysnoob

I will attach that page

12. modphysnoob

13. prakharluv

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

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

how did they come to that assumption?

16. modphysnoob

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

17. Jemurray3

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

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

19. modphysnoob

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

mhmm

21. modphysnoob

not quite?