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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1The induced emf in the circuit will be (ignoring the minus sign for the moment) \[ EMF = \frac{d}{dt} \Phi_B = \frac{d}{dt} B\cdot A = B \cdot L \cdot \frac{dx}{dt} = BLv\]

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1dw:1362247929761:dw

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1But \[EMF = BLv = IR \] so the current \[ I = \frac{BLv}{R} \]

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1No, not really. What book is this?

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1I am unfamiliar with it, sorry. I don't know, unless there's more to the question.

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

prakharluv
 one year ago
Best ResponseYou've already chosen the best response.0ok 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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1Oh, I see. What they are saying is the magnetic force on the particles is the same as if there was an electric field \[\vec{E} = (\vec{v}\times \vec{B}) \]

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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.1yes. 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 \[\vec{E} = \vec{v}\times \vec{B} \]

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

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