Question

I've a huge question about a problem of Magnetism pls pls help (I'll add the drawing)

Answer 1

The square is moving at V=Vo in y

Answer 2

I need the EMF

Answer 3

http://www.howstuffworks.com/magnetism-info.htm
http://www.ducksters.com/science/magnetism.php
Hope this helps :) I learned this at some point this year i just dont remember... but these links might help

Answer 4

Not really, I get how it works my problem is that I should get to an expresion \[EMF=\frac{ Vo*a^{2} *\mu_{o}*I}{ 2*\pi*p*(p+a) }\]

Answer 5

O_O over my head.. sorry :/

Answer 6

A square loop of A side moves away in direction Y from infinite length filament with current I along z. Assuming that P = Po at the time t = 0 shows that EMF when t> 0 :EMF=Vo∗a2∗μo∗I2∗π∗p∗(p+a)

Answer 7

ok i think i get it

first thing is that the magnetic field lines are into the book.. so its perpendicular to the square loop and hence flux lines are nomal

emf = rate of change of flux

so lets first find flux at any time t

let it be at a distance of x from the current carrying wire..

then the flux \[\phi_b = \int\limits_{x}^{x+a}B.dl \]
m sure you can calulate that integral..

so you ll get phi as a function of x
so when you differentiate you ll have a dx/dt term on the right side.. that would be v

and using initial conditions when t = 0 x = p0 should give you the final required expression.. m lazy to do the derivation myself :P

Answer 8

Here is my proposal for the solution. Any doubt, please ask