Here's the question you clicked on:
modphysnoob
EMF
|dw:1362247626272:dw|
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
is there are any dimensions of loop?
The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]
|dw:1362247929761:dw|
But \[EMF = BLv = IR \] so the current \[ I = \frac{BLv}{R} \]
I see, so there is no point to be messing around the force at all?
No, not really. What book is this?
Fundamental of Applied Electromagnetics by ulaby
I am unfamiliar with it, sorry. I don't know, unless there's more to the question.
I will attach that page
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
Oh, I see. What they are saying is the magnetic force on the particles is the same as if there was an electric field \[\vec{E} = (\vec{v}\times \vec{B}) \]
how did they come to that assumption?
so magnetic field here would push all positive particle down , negative particle up , between them is potential difference
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 \[\vec{E} = \vec{v}\times \vec{B} \]
I understand the formula , the part that confused me is why these two force must be equal
oh, I think I understand , it is one force , not two force; we are just looking at it as if it two way