# Faraday Lenz lab - My calculations are Way off and I'm not sure why

1. Mar 27, 2014

### uterii

The purpose of the lab was to measure, and also calculate, the induced current in a disconnected coil, due to a second coil connected to a power supply. Involved magnetic fields and electric fields (I think.

For the purposes of this lab, the magnetic permittivity is 1.26*10-6, the number of turns is 500, and the radius of the coil is 10.5cm.

B=N (μ_0 I)/2R
ϕ=B∙A
=N*(u0*I)/2 * R2 /(R2 +x2 )3/2

Magnetic Field Created by the First Coil
I_left=18Ω/18V=1 Amp
B_left=500*((1.26*〖10〗^(-6) )*1Amp)/2*(.105m)^2/〖〖〖((.105m)〗^2+(.105m)〗^2)〗^(3/2) = 0.0010 Tesla
Current in Second Coil when Voltage in First Coil is Constant
In this situation, there is no induced current in the second loop. For an induced current to exist, there must be a change in flux over time. This is impossible when current is held constant.
Current in Second Coil when Voltage Linearly Increased over 10sec
ϕ_initial=0 Wb
ϕ_final= 0.0010 Tesla*(π*(10.5cm)^2 )=3.464*〖10〗^(-5) Wb
ε=-((3.464*〖10〗^(-5) Wb-0 Wb))/10s=3.46*〖10〗^(-6) V
I_induced=18Ω/(3.46*〖10〗^(-6) V)=5196896 Amp

My TA has told us that the induced current should be calculated as ~0.1mA, and experimentally the induced current was found to be 0.6mA. Obviously I am WAY off, but I'm honestly not sure what I'm doing wrong.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Mar 27, 2014

### BvU

I_induced=18Ω/(3.46*〖10〗^(-6) V)=5196896 Amp is weird. Wasn't Ohms law a little different ?

Any measurement results for the induced voltage ?
I must say I find a change of 1 A in 10s rather slow, so maybe these low values can be expected.
I do find the $\Phi_{\rm final}$ value you found rather low. Can you check ?

Is there a drawing of the setup ? How far apart are the two coils ? Orientation ?

3. Mar 27, 2014

### uterii

The two coils are 10.5cm apart, and they are paralell to one another. Each has a radius of 10.5cm, and the resistance of each coil is 18Ω.

We did not measure the induced voltage, just the induced current.

Ohm's law is V=IR. Ah, I see a mistake! So that was part of my problem, but something before that step is still off.

I've checked my calculation of $\Phi_{\rm final}$, and I'm getting the same answer, so I'm worried it is an algebra error.

When inducing a current into the second coil, we had the power source increase linearly from 0V to 18V in 10seconds

Last edited: Mar 27, 2014
4. Mar 27, 2014

### BvU

Good. Now I use your first formula and get B=N (μ_0 I)/(2R) and get 0.003 T.

What do you do to get $\Phi$ ?

Do you now see why the rules and guidelines force using the template ? Saves time, yours and ours.... Gets you better help too.