How Does Changing Electric Field Influence Magnetic Field in a Capacitor?

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Homework Help Overview

The discussion revolves around a parallel-plate capacitor with circular plates, focusing on how a changing electric field influences the magnetic field between the plates. The original poster attempts to calculate the magnetic field at a specific point within the capacitor using Ampere's Law and related equations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Ampere's Law and the relationship between the changing electric field and the induced magnetic field. There are attempts to verify calculations and clarify the variables involved, particularly the distinction between dV/dt and dE/dt.

Discussion Status

Some participants have provided calculations and questioned the correctness of their approaches. There is an acknowledgment of confusion regarding the variables used in the equations, and a few participants suggest revisiting the original problem statement for clarity. Guidance has been offered regarding the correct interpretation of the changing electric field.

Contextual Notes

Participants note the importance of correctly identifying the rate of change of the electric field versus the voltage change, which has implications for the calculations being discussed. There is also mention of the speed of light in vacuum as a relevant constant in the context of the problem.

GeorgeCostanz
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Homework Statement



A parallel-plate capacitor has circular plates with radius 49.0 cm and spacing 2.20 mm. A uniform electric field between the plates is changing at the rate of 1.90 MV/m/s. Find the magnitude of the magnetic field between the plates at a point 12.1 cm from the axis (entirely inside the capacitor).

Homework Equations



Ampere's Law for induced current

B(2∏r) = (μ-naught)(ε-naught)(A/d)(dV/dt)

r = .121m
μ-naught = 4∏x10^-7
ε-naught = 8.85x10&-12
d = .0022m
dV/dt = 1.9x10^6 V/(m/s)
A = ∏(.49m)^2

The Attempt at a Solution



the answer is B = 1.28x10^-12 T, but i can't seem to get that answer using my equation. i'd appreciate it if someone could direct me toward my error

thanks
 
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Hi George! :smile:

(try using the X2 button just above the Reply box :wink:)

Show us your full calculations.

(In particular, what did you get for the current through the cylinder of radius 0.121 m ?)
 
@tiny-tim

sure.

Id = [ (E\circA)/d ] * (dV/dt)

Id = [ (E\circ*(∏(.492))/.0022 ] * (1.9x106)

Id = .005765 A

i guess :smile:
 
GeorgeCostanz said:
B(2∏r) = (μ-naught)(ε-naught)(A/d)(dV/dt)

mmm … your formula seems to be correct, but I'm not getting the result of 1.28 10-12 T either :confused:
 
i got the right answer using the following equation (googled the question)

B = [ (1/2)(r)(dV/dt) ] / C2

C = 3x108 = speed of light in vacuum
dV/dt = 1.9x106
r = .121m

not sure how the 2 equations are related tho
 
ah! :rolleyes: i took your word for it instead of looking at the original question …
GeorgeCostanz said:
A uniform electric field between the plates is changing at the rate of 1.90 MV/m/s.

dV/dt = 1.9x10^6 V/(m/s)

noooo :redface: … that wasn't dV/dt, it was dE/dt ! :biggrin:

is everything clear now? :smile:

(and c2 = 1/µoεo, which btw would have have been a lot easier for you to use :wink:)
 
tiny-tim said:
ah! :rolleyes: i took your word for it instead of looking at the original question …


noooo :redface: … that wasn't dV/dt, it was dE/dt ! :biggrin:

is everything clear now? :smile:

(and c2 = 1/µoεo, which btw would have have been a lot easier for you to use :wink:)


wow I'm slow
it's a miracle I've even made it this far

so for r < R,

B = (1/2)µoεo(r)(dE/dt)

thus B = [ (1/2)(r)(dE/dt) ] / C2

guess i got mixed up with all these equations/derivations in front of me

thanks tiny-tim
 

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