- #1
uterii
- 2
- 0
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.
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.