Question

Greetings, I would like to know what the actual infinitisimal magnetic flux (dΦ) of a conductive ring is. More details below.

Answer 1

* I ask for the infinitesimal flux of the conductive ring, and after that the EMF

* B =Cxxˆ−Cyyˆ
* V is constant on the x axis (we have only X axis velocity)

* (0,0) center of the ring

My results: for dφ I found this to be like Cx2πadx and EMF -2πa(v^2)t

Answer 2

* note that dΦ = \[\vec{B} \vec{dS}\] or \[\int\limits_{strip}^{}\vec{B} d\vec{A}\] where strip is the orthogon strip that connects the ring in thr two phases (before and after)

Answer 3

Is axis of the ring parallel to x-axis?

Answer 4

the axis of the ring is the x-axis - (0,0) centered ring

Answer 5

is the normal area vector ring perpendicular to B or parallel??

Answer 6

if perpendicular then 0 else it will depend on the order of integration..

Answer 7

wait check the B field above, it is neither perpendicular nor parallel

Answer 8

is there one or two rings??

Answer 9

one ring during an infitesimal motion at x-axis, let's say this to be dx

Answer 10

x axis is its centroidal axis?? sorry cant see where is the disk moving..

Answer 11

which products a dΦ οf \[\vec{B} \vec{dS}\] or \[\int\limits_{strip}^{}\vec{B}\vec{dA}\]

Answer 12

centroidal

Answer 13

**where strip is the infitesimal orthogonal strip with sides dl=dx and dl=adθ

don't forget that during infitesimal motions we ignore the Φ of the main surfaces because thry are the same and can't be included in dΦ so we have only to do with side flux

Answer 14

(**i opened this topic just for discussion and to check my answers to this problem)

Answer 15

i cant see what the fig is like..but project the ring on XY plane and then apply the formula by taking a stripparallel to x axis..

Answer 16

just imagine 2 rings on almost the same location (distance=dx) and that "dx line" connecting them.. then imagine the surface they create (a cylinder surface)

Answer 17

also a second way to view this according to my opinion is to consider a small part of this "cylinder" which makes a ring again. Then create very small rectangular parts on it's surface with side 1 = dx and side 2 = dl (but dl is a polar motion due to the electric current which advances on the ring) so side 2 = dl = adθ

Answer 18

if anyone disagrees just let me know

Answer 19

Magnetic flux through ring is \(\Phi=\pi a^2Cx\) and electromotive force is \(e=-\pi a^2Cv\)
where \(a\) is the radius of the ring.

I do not find wise to calculate the magnetic flux 'cut' by the ring.
you will need to integrate \(\cos ^2\theta \:d\theta\)

Answer 20

yes but we need dΦ not Φ, and it comes up to a linear flux. Check Griffiths Introduction to Electrodynamics of Faraday's Law/Magnetic Flux part, they say that we can work like this to find the dΦ

Answer 21

S (main) surface is stable so the dΦ comes from linear flux which comes from infinitesimal rectangulars

Answer 22

Of course you can, but it is much more complicated, than working the actual flux through the circuit.

You will get a similar expression: \(d\Phi=-\pi a^2C dx\)

Answer 23

Thank you for for reply Vincent.

According to Griffith's Introduction to Electrodynamics, the dΦ comes from the following integral: Φstripe where stripe is the "line" that sticks the two phases together. Imagine something like the following (IMPORTANT: the field is not homogeneous)





this shape represents the magnetic field and the infimitesial motion of the ring (**continue bellow)