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modphysnoobBest ResponseYou've already chosen the best response.0
dw:1362247626272:dw
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.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
 one year ago

prakharluvBest ResponseYou've already chosen the best response.0
is there are any dimensions of loop?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
dw:1362247929761:dw
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
But \[EMF = BLv = IR \] so the current \[ I = \frac{BLv}{R} \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
I see, so there is no point to be messing around the force at all?
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
No, not really. What book is this?
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
Fundamental of Applied Electromagnetics by ulaby
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.1
I am unfamiliar with it, sorry. I don't know, unless there's more to the question.
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
I will attach that page
 one year ago

prakharluvBest ResponseYou've already chosen the best response.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
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.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 \[\vec{E} = (\vec{v}\times \vec{B}) \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
how did they come to that assumption?
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
so magnetic field here would push all positive particle down , negative particle up , between them is potential difference
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.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 \[\vec{E} = \vec{v}\times \vec{B} \]
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
I understand the formula , the part that confused me is why these two force must be equal
 one year ago

modphysnoobBest ResponseYou've already chosen the best response.0
oh, I think I understand , it is one force , not two force; we are just looking at it as if it two way
 one year ago
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