- #1
JustinLevy
- 895
- 1
Reading a different homework problem here made me realize I don't understand somethings about inductors. I made a simple problem that focuses just on the inductor to help me learn. Please do help me solve this problem.
Consider this:
The loop has inductance L, no resistance, and that initially there is no current in the loop. Then at time t=0, a current source is used to increase the current at a steady rate (I = a t).
Current goes up the green wire, around the blue loop, and then up the red wire. Additionally, consider a voltmeter connected with the ground on the red point, and the "positive" terminal on the green point.
What does the volt meter read?
I'm not sure what you want here.
I believe a voltmeter reads (with leads on points 'a' and 'b'):
[tex] Voltage = - \int_a^b \mathbf{E} \cdot d\mathbf{l} [/tex]All of electrodynamics is in these five equations:
[tex] \nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}[/tex]
[tex] \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t} [/tex]
[tex] \nabla \cdot \mathbf{B} = 0 [/tex]
[tex] \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} [/tex]
[tex] \mathbf{F} = q (\mathbf{E} + \mathbf{v}\times\mathbf{B})[/tex]
The definition of flux, and inductance:
[tex] \Phi = \int_S \mathbf{B} \cdot d\mathbf{a}[/tex]
[tex] \Phi = L I [/tex]
In case it is useful, there is also the definition of the potentials:
[tex] \mathbf{E} = - \nabla V - \frac{\partial}{\partial t} \mathbf{A}[/tex]
[tex] \mathbf{B} = \nabla \times \mathbf{A}[/tex]
We will use Coloumb gauge here:
[tex] \nabla \cdot \mathbf{A} = 0[/tex]
There are a few steps in my logic, please let me know if they are correct.
part 1]
Using Maxwell's equations, when the current starts increasing a magnetic field will be generated in the upward direction inside the loop in that diagram. In other words, the magnetic flux of the loop will increase with time (with 'up' defined as positive).
Correct?
part 2]
Due to the negative sign in Del.E = - (d/dt) B, an electric field will be generated to oppose this change in flux (this is also known as Lenz's law). So the electric field will be pointing against the current around the loop. Correct?
part 3]
Similarly, this means the integral of - E.dl from the red to the green point is negative.
This seems to argue the voltmeter would read negative.
Correct?
If I actually plug in all the math along these steps, I get:
Voltage = - L dI/dt = - L aHowever, I know that this is wrong, as the answer I should be getting is:
Voltage = L dI/dt = L a
and furthermore, if it actually was negative, then the current would be flowing against the electric field which doesn't make any sense.
What the heck is going on here?
Which part of logic above is incorrect?
Homework Statement
Consider this:
The loop has inductance L, no resistance, and that initially there is no current in the loop. Then at time t=0, a current source is used to increase the current at a steady rate (I = a t).
Current goes up the green wire, around the blue loop, and then up the red wire. Additionally, consider a voltmeter connected with the ground on the red point, and the "positive" terminal on the green point.
What does the volt meter read?
Homework Equations
I'm not sure what you want here.
I believe a voltmeter reads (with leads on points 'a' and 'b'):
[tex] Voltage = - \int_a^b \mathbf{E} \cdot d\mathbf{l} [/tex]All of electrodynamics is in these five equations:
[tex] \nabla \cdot \mathbf{E} = \frac {\rho} {\varepsilon_0}[/tex]
[tex] \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t} [/tex]
[tex] \nabla \cdot \mathbf{B} = 0 [/tex]
[tex] \nabla \times \mathbf{B} = \mu_0\mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}} {\partial t} [/tex]
[tex] \mathbf{F} = q (\mathbf{E} + \mathbf{v}\times\mathbf{B})[/tex]
The definition of flux, and inductance:
[tex] \Phi = \int_S \mathbf{B} \cdot d\mathbf{a}[/tex]
[tex] \Phi = L I [/tex]
In case it is useful, there is also the definition of the potentials:
[tex] \mathbf{E} = - \nabla V - \frac{\partial}{\partial t} \mathbf{A}[/tex]
[tex] \mathbf{B} = \nabla \times \mathbf{A}[/tex]
We will use Coloumb gauge here:
[tex] \nabla \cdot \mathbf{A} = 0[/tex]
The Attempt at a Solution
There are a few steps in my logic, please let me know if they are correct.
part 1]
Using Maxwell's equations, when the current starts increasing a magnetic field will be generated in the upward direction inside the loop in that diagram. In other words, the magnetic flux of the loop will increase with time (with 'up' defined as positive).
Correct?
part 2]
Due to the negative sign in Del.E = - (d/dt) B, an electric field will be generated to oppose this change in flux (this is also known as Lenz's law). So the electric field will be pointing against the current around the loop. Correct?
part 3]
Similarly, this means the integral of - E.dl from the red to the green point is negative.
This seems to argue the voltmeter would read negative.
Correct?
If I actually plug in all the math along these steps, I get:
Voltage = - L dI/dt = - L aHowever, I know that this is wrong, as the answer I should be getting is:
Voltage = L dI/dt = L a
and furthermore, if it actually was negative, then the current would be flowing against the electric field which doesn't make any sense.
What the heck is going on here?
Which part of logic above is incorrect?