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modphysnoob
 2 years ago
EMF
modphysnoob
 2 years ago
EMF

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

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

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

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

modphysnoob
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1No, not really. What book is this?

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

Jemurray3
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0I will attach that page

prakharluv
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0how did they come to that assumption?

modphysnoob
 2 years 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
 2 years 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
 2 years 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
 2 years 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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