Magnitude of induced voltage in conductor within AC solenoid

[uby]

Please note, I'm not in school anymore -- just a basic physics question!

Homework Statement

A conductor is placed inside the coil of an induction heater.

Imagine that the inductor coil has radius R, number of turns N, operating frequency f [Hz], drawing current I = I0 * sin(2*pi*f*t).

The conductor is a cylinder of radius r located concentric to the inductor.

Calculate the magnitude of the induced voltage V at a fixed position on the outer surface of the conductor as a function of time.

Homework Equations

V = dO/dt = d(B*A)/dt (where O is the magnetic flux given by B dot A, assuming entire flux goes through cross-sectional area)

B = mu * I / 2*R (where mu is permeability of air = (4*pi)E-7 T*A/m)

The Attempt at a Solution

V = d(B*A)/dt = d(mu*I*A/2*R)/dt = d(mu*I0*sin(2*pi*f*t)*A/2*R)/dt
V = mu*I0*A/2*R * d(sin(2*pi*f*t))/dt
V = pi*f*mu*I0*A/R *cos(2*pi*f*t)

A = pi*(R^2-r^2) is the area where magnetic field flux operates on conductor

so, for any position on the outer cylinder radius, the maximum magnitude of the induced voltage is found to be:
max(V) = pi^2*(R^2-r^2)*f*mu*I0/R

putting some rough numbers on this:

let R = 0.05 m, r = 0.025 m, f = 100000 Hz, I0 = 20 Amps,
then max(V) = 0.93 VDoes this look correct?

About 1V may not seem like much, but it is for my intended application! And if I need to use megahertz range frequencies the induced voltage goes up by orders of magnitude!

Thanks!

 

[member 553083]

Yes, your calculation looks correct. The maximum magnitude of the induced voltage is dependent on the parameters you've specified, so if any of these are changed, the result may change as well.