Determine the maximum magnetic flux through an inductor

[gnome]

I'm sure I'm going to 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 (ΔV_rms = 120 V., f = 60 Hz.)

That's all that's given.

I know that ΔV_max = 120*√2 and ω = 2Πf = 120Π and I think the magnetic flux will be at a maximum when the current is at a maximum, and
I_max = (ΔV_max) / (ω L) so

I_max = (120*√2) /(120ΠL)

Also, Φ_B = I*L/N so I get

Φ_B = (120*√2) /(120Π N)

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 )

 

[arcnets]

gnome,
I think \Phi_B=IL, see here:

http://scienceworld.wolfram.com/physics/Inductance.html

 

[gnome]

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

As you found in that Wolfram link, it comes from Faraday's
ε = -dΦ_B/dt
and
dΦ_B/dt ~ dI/dt (using ~ to mean proportional to)

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

dΦ_B/dt = L*dI/dt

and then, by integrating we get

Φ_B = L* I

My objection is that in a coil with N turns, the inductance is
L = (NΦ_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 * Φ_B, which of course is just
(120*√2) /(120Π) = .450 T*m²

 

[arcnets]

Did you kick yourself?