Homework Help: Finding RMS current in an AC solenoid

1. Aug 2, 2011

chinye11

1. The problem statement, all variables and given/known data
A 2000 turn solenoid of length 1.50 m and diameter 5.00 cm has a dc resistance of 3.0. It is
connected to a 50 Hz, 40.0 Volt (rms) ac supply. Find the rms current in the solenoid

2. Relevant equations
Vind = L di/dt
Vind = Induced EMF
Ohm's Law
V-Vind = IR
L = (4pi x 10^-7)n^2 Al

Where L=self inductance and l = length

3. The attempt at a solution
This is my first time using this website and i am not sure how to get in symbols, I will do my best to be clear

I started finding the inductance.

I then used Ohm's Law: V-Vind=IR

After mathematical manipulation I got I = V/R[1-e^(Rt/L)]

My question is as regards the frequency. It seems to have no effect on the system if you use my method yet if I try to find the induced E.M.F. using Faraday's Law on the magnetic flux you will find:

Vind= 0.05 Sin(theta) I(t)
Vind= 0.05 Sin (2 pi (f) t) I(t) from theta=wt

So I was wondering if someone could explain both the concept and the calculations of the effect of the frequency. Thank You.

2. Aug 2, 2011

chinye11

Just to clarify I am aware that the current is periodic and that the frequency can help in calculating max current and other such things however, I was wondering if it was of any use in this specific question.

3. Aug 2, 2011

Staff: Mentor

You've effectively got a resistor and inductor in series, and it's driven by a 50Hz AC supply. If you were to calculate the effective impedance (Z) of the resistor+inductor at the given frequency, you could apply Ohm's law to find the current: I = V/Z. The magnitude of I is what you're after.

For an inductor the impedance varies in direct proportion to the frequency.

This presumes that you've been introduced to the concept of complex impedance...

4. Aug 2, 2011

chinye11

Thanks very much I hadnt covered impedance well. I have since gone through and understand it.