anonymous
  • anonymous
EMF
Physics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
|dw:1362247626272:dw|
anonymous
  • anonymous
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
prakharluv
  • prakharluv
is there are any dimensions of loop?

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anonymous
  • anonymous
The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]
anonymous
  • anonymous
|dw:1362247929761:dw|
anonymous
  • anonymous
But \[EMF = BLv = IR \] so the current \[ I = \frac{BLv}{R} \]
anonymous
  • anonymous
I see, so there is no point to be messing around the force at all?
anonymous
  • anonymous
No, not really. What book is this?
anonymous
  • anonymous
Fundamental of Applied Electromagnetics by ulaby
anonymous
  • anonymous
I am unfamiliar with it, sorry. I don't know, unless there's more to the question.
anonymous
  • anonymous
I will attach that page
anonymous
  • anonymous
prakharluv
  • 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
anonymous
  • anonymous
Oh, I see. What they are saying is the magnetic force on the particles is the same as if there was an electric field \[\vec{E} = (\vec{v}\times \vec{B}) \]
anonymous
  • anonymous
how did they come to that assumption?
anonymous
  • anonymous
so magnetic field here would push all positive particle down , negative particle up , between them is potential difference
anonymous
  • anonymous
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 \[\vec{E} = \vec{v}\times \vec{B} \]
anonymous
  • anonymous
I understand the formula , the part that confused me is why these two force must be equal
anonymous
  • anonymous
oh, I think I understand , it is one force , not two force; we are just looking at it as if it two way
anonymous
  • anonymous
mhmm
anonymous
  • anonymous
not quite?

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