How Does Discharging a Capacitor Affect Its Internal Magnetic Field?

[Ben Whelan]

Homework Statement

A cylindrical parallel plate capacitor of radius R is discharged by an external current I. The
total electric field flux inside the capacitor changes at a rate dΦ_e/dt = I/ ε₀. What is the
strength of the resulting magnetic field B(r, I) inside the capacitor at a radial distance r
from the centre axis? Start the answer with Maxwell’s extension to Ampere’s law.
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Homework Equations

So using the line integral ∫B⋅dl= μ₀ (I + I_d )[/B]

The Attempt at a Solution

you get B⋅2πr= μ₀ (I + I_d )

However i don't understand what I and I_d are?

I have the equation I_d= ε₀ dΦ_e/dt

however i don't understand why this is true or where this takes me?
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[rude man]

You have an equation and you don't know what the symbols represent? May I suggest your first step is to find that out.
(Hint: the problem tells you what I is; it's given.)

 

[Ben Whelan]

You have an equation and you don't know what the symbols represent? May I suggest your first step is to find that out.
(Hint: the problem tells you what I is; it's given.)

Sorry i should have qualified, I know what the symbols represent, and i realize that I_d is given, i also assume I is 0. However what i don't understand is why?

 

[rude man]

Sorry i should have qualified, I know what the symbols represent, and i realize that I_d is given, i also assume I is 0. However what i don't understand is why?

Actually, I is given, not I_d. I is the current in the wiring, not inside the capacitor. Inside the capacitor, I = 0 as you say.
Your answer is to be B(I,r).
Do you know the Maxwellian extension of Ampere's law? It relates the circulation of B to the electric flux inside the circulation perimeter.
Hint: the perimeter is itself a function of r. It is inside the circular plates.