Calculating Electromotive Force in a Time-Dependent Magnetic Field

[Frank Einstein]

Homework Statement

I have a problem with the next situation. I have a magnetic dipole moving at constant speed on the z axis. On the plane z=0 I have a circular wire with a resistance R and radius a. I have to calculate the electromotive force on the wire as a function of the speed of the dipole, v, the position of the dipole ,z, and the dipolar moment m.

Homework Equations

ε=-dΦ/dt, where ε is the electromotive force and Φ is the magnetic flux.
B=(μ₀/(4π))*(3r(r*m)/r⁵-m/r³)
Φ=∫Bds
ds=( cosφ, sinφ, 0)
r = (a cos φ, a sinφ, z)

The Attempt at a Solution

Once I calculate a huge expression for the magnetostatic field, B, and I integrate to find the flux, my problem comes; I have to make a time derivate, and the only time dependent factor here is z; but Φ(r(z(t))) do I have to derivate (∂Φ/∂z)(dz/dt)=(∂Φ/∂z)v or do I have to do something else?

 

[BiGyElLoWhAt]

Why are you trying to calculate the magnetostatic field? Shouldn't the field be dynamic? Otherwise there would be no emf.

I don't see a way around integrating spatially and then differentiating temporally, if that's what you're asking. Also, you're using cartesian coordinates? I think your life would be much easier if you worked in cylindrical. Just my opinion though.

 

[Frank Einstein]

Well I Integrate in cartesian because I feel more comfortable doing operating with them; what is more, when I make the scalar product BdS I have an scalar multipied by sin²+cos², so the dependence with the angle dissapears. My main concern here is not about the coordinates, but how to determine the electromagnetic foerce, -d/dt(∫BdS)

 

[BiGyElLoWhAt]

One of the variables in B is a function of time.

 

[Frank Einstein]

yes, B(r(z(t))), there is whem my doubt comes, do I simply derivate ∂B/∂z and them muptiply by dz/dt=v?

 

[BiGyElLoWhAt]

Does r(t) equal z(t) for all points on the surface?

 

[Frank Einstein]

No, r=Sqrt(z^2+a^2), where a is the radius of the wire

 

[BiGyElLoWhAt]

Ok, so do you know how to chainrule derivatives? You either have to chainrule twice or substitute, frankly.

 

[Frank Einstein]

That's it, I only wanted to know if I had to apply the chain rule twice.
Thanks.

 

[BiGyElLoWhAt]

No problem

 

[Feodalherren]

I realize you already solved it but I strongly suggest that you look up a youtube video on cylindrical/spherical coordinates. It makes E&M so much easier and if you ignore the derivation there are only two very easy formulas to remember.

Just my two cents. It will be well worth a 30 min or-so investment of your time.