# Magnetic flux

1. Nov 30, 2003

### gnome

I'm sure I'm gonna kick myself when someone shows me the way, but there's a limit to how long I can stare at this question...

Determine the maximum magnetic flux through an inductor connected to a standard outlet (&Delta;Vrms = 120 V., f = 60 Hz.)

That's all that's given.

I know that &Delta;Vmax = 120*&radic;2 and &omega; = 2&Pi;f = 120&Pi; and I think the magnetic flux will be at a maximum when the current is at a maximum, and
Imax = (&Delta;Vmax) / (&omega; L) so

Also, &Phi;B = I*L/N so I get

but I have no idea what the value of N is for the inductor, & I can't see any way to get rid of that N.

(edited to correct 60 x 2 = 120, not 100 )

Last edited: Dec 1, 2003
2. Dec 1, 2003

3. Dec 1, 2003

### gnome

Yeah, thanks Arcnets, that's basically the answer. My objection is that &Phi;B = IL if there's only one turn.

&epsilon; = -d&Phi;B/dt
and
d&Phi;B/dt ~ dI/dt (using ~ to mean proportional to)

From that, the inductance L is defined as the proportionality constant, so

d&Phi;B/dt = L*dI/dt

and then, by integrating we get

&Phi;B = L* I

My objection is that in a coil with N turns, the inductance is
L = (N&Phi;B)/I
and any inductor in a circuit is going to have more than 1 turn, so it seems to me that the question was, to say the least, ambiguous.

But, I asked the professor about it today and, yes, his answer was that the question "meant" the total flux through all the turns, in other words, they were looking for the value of N * &Phi;B, which of course is just
(120*√2) /(120Π) = .450 T*m2

4. Dec 2, 2003

### arcnets

Did you kick yourself?