Power generated by a coil rotating in magnetic field

  • #1
9
1
The problem: a coil of radius r, length l and N turns, rotating with constant angular velocity ω around an axis perpendicular to its simmetric axis and passing for the center of the coil. The coils is submersed in a static magnetic field, intensity B0, perpendicular to the axis of rotation of the coil. The coil is connected to a resistance R. What is the medium power dissipated by resistance?
1st solution: solving a differential equation with the current i(t) unknow I obtain:
$$-\frac{\text{d}}{\text{d}y}(B_0N\pi r^2cos(\omega t)+\frac{\mu_0 N^2 \pi r^2 i(t)}{l})=Ri(t)$$
and then the current:
$$i(t)=\frac{Rl^2 N \pi r^2\omega B_0}{((Rl)^2+(\mu_0 N^2 \pi r^2\omega)^2)^2}sin(\omega t)-\frac{\mu_0 N^3(\pi r^2)^2 \omega^2 B_0 l}{((Rl)^2+(\mu_0 N^2 \pi r^2\omega)^2)^2}cos(\omega t)$$
and medium power P:
$$P=\frac{R}{2} \frac{N^2 (\pi r^2)^2 \omega^2 B_0^2 l^2}{(Rl)^2+(\mu_0 N^2 \pi r^2\omega)^2}$$
2nd solution: a virtual ca generator in series with the coil. The circuit impedance is:
$$Z=R+j\omega L$$
the Faraday-Newman-Lentz applied to the solenoid get the max fem:
$$V= N \pi r^2 \omega B_0 sin(\omega t) $$
and the current max:
$$I=\frac{V}{|Z|}$$
so, the medium power dissipated by resistance:
$$P=\frac{R}{2} I^2=\frac{R}{2} \frac{(N \pi r^2 \omega B_0)^2}{R^2+(\omega L)^2}$$
and, substituting ##L=\frac{\mu_0 N^2\pi r^2}{l}##, I obtain the same than 1st solution
3rd soultion: somewhere in internet you can find:
the Faraday-Newman-Lentz applied to the solenoid get the max fem:
$$V= N \pi r^2 \omega B_0 sin(\omega t) $$
so the max current on the resistor:
$$I=\frac{V}{R}$$
and, then the medium power:
$$P=\frac{R}{2} VI=\frac{R}{2} \frac{(N \pi r^2 \omega B_0)^2}{R^2}$$
The question:
1) have you evidence of this calculation?
2) do you know a textbook for this problem?
3) do you know a software to simulate this system (matlab, pspice, labview, ...)?
thamks!
 
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Answers and Replies

  • #2
osilmag
Gold Member
136
32
I have never seen that before. For a coil moving through a magnetic field (and vice versa), I think Faraday's Law would be useful to find the EMF and thus the current.
 
  • #3
442
119
If the coil data meets the attached sketch then
N*π*r^2*ω*Bo*sin(ω*t)+μo*N^2*πr^2/l *di(t)/dt+R*i(t)=0it seems- to me- it is correct if the core is of magnetic material made [it is not an air core].
The general solution it has to be like i(t)=I*sin(ω*t+φ)+Io*e^(-t/T)+Const
T=-R/(μo*N^2*r^2/l)
If t≈∞ then i(t)=I*[sin(ω*t)*cos(φ)+cos(ω*t)*sin(φ)]
The second solution it is better I think.
upload_2019-1-6_8-40-50.png
 

Attachments

  • #4
rude man
Homework Helper
Insights Author
Gold Member
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I don't think thed length of the coil has anything to do with it.
 

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