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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?

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The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]
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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
mhmm
not quite?

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