Heat dissipated by a cylinder in a magnetic field

[Mr.Joule]

Homework Statement

We have a cylinder (radius r, height h, conductivity γ) and it is immersed in a variable magnetic field B(t) = B0* e^(-t/T). The field is parallel to the axis of the cylinder.

The problem then asks to find the heat dissipated by the cylinder after an infinitely long time.

Homework Equations

W = \int R i^2 dt

i = (-1/R) * \pi r^2 dB/dt

The Attempt at a Solution

I consider the cylinder as a succession of circular loops.

Let's consider one loop:

The heat dissipated is: W = \int R i^2 dt

where

R = \rho \frac{l}{S} = \frac{2\pi r}{γ*S}

i^2 = (E/R)^2 = (1/R^2) (\frac{d\varphi(B)}{dt})^2 = (1/R^2) \pi^2 r^4 (dB/dt)^2

where E is the Electromotive force.

So:

(dB/dt)^2 = (B0/T)^2 * e ^ {-2t/T}

W = \int R i^2 dt = R/R^2 \pi^2 r^4 (B0/T)^2 \int_0^\infty e^{-2t/T} dt =(\pi^2 r^4 (B0/T)^2)*T/2R

Since there are h loops, the result is \frac{h*\pi^2 *r^4 (B_0)^2}{2RT}

Is this right?

Thank you in advance.

 

[rude man]

You're on the right track, but realize that h → ∞ !
And so of course the current di in each loop must be infitesmally small, also that the resistance of a loop will vary with r, 0 < r < radius. (I am making r variable. Dumb to use r for radius, but can't use R either since R is resistance).

In other words, consider an annulus of differential width dr and get the corresponding resistance R(r), then di = emf(r)/R(r), then integrate dP = (di^2)*R from 0 to r, then integrate dP over time t from 0 to ∞ (energy = ∫power*dt).

Good problem!

 

[Mr.Joule]

Hey :) Thanks a lot for your help.

Let's give it a second try step by step.

1. why does h → ∞ ? it isn't an infinitely long cylinder!

2. The cross section of the wire is A (constant), so

dR(r) = rho * dL/A = (rho * 2pi * dr)/A, right?

Now, di = emf(r)/R(r) with

d emf(r) = 2pi*rdr dB/dt

but now di = (dB/dt)*A*r since dr disappeared. I think it can't be right. :\

 

[rude man]

Hey :) Thanks a lot for your help.

Let's give it a second try step by step.

1. why does h → ∞ ? it isn't an infinitely long cylinder!

Mea culpa. I got confused when you spoke of "h loops".

OK, think of an elemental annulus, height = h, thickness = dr, radius = r. The annuli comprising the cylinder range in radius fro 0 to R. (I will call italicized R the radius of your wire/cylinder.) In other words, think of the cylinder as kind of a cylindrical onion, with the annuli represented by the onion's layers.

There will be an emf generated around this elemental annulus. The conductance of this annulus will be dS = γ*h*dr/2πr Siemens. You need to visualize the annulus as being unfolded to a slab of metal with dimensions length = 2πr width = h, thickness = dr.

The emf around this annulus' contour will be area * dB/dt (ignoring sign). Area = πr².

The current di will then be emf*dS, and the power dissipation will be dP = (emf)²*dS.

To get the total power you now need to integrate dP from r = 0 to r = R.

Finally you need to integrate P from t = 0 to t = ∞ to get the total energy, which of course is the heat you're looking for.





2. The cross section of the wire is A (constant), so

dR(r) = rho * dL/A = (rho * 2pi * dr)/A, right?

Now, di = emf(r)/R(r) with

d emf(r) = 2pi*rdr dB/dt

but now di = (dB/dt)*A*r since dr disappeared. I think it can't be right. :\[/QUOTE]

The current does not flow along the wire. The current flows in circles inside the wire. Other than that I don't follow your math. Where did 2pi*dr come from?