Current of a loop as a function of time

By Vamsi Challa

Introduction

The purpose of this derivation is to relate the current of a loop of wire as a function of time. Imagine a loop of wire in between a capacitor. As the battery begins to charge up, how does the current of the wire relate to time and other important variables within the circuit?

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Finding the Electric Field

In order to find the electric field, we will use Gauss’s Law. ∫EcosθdA = q/ε. This relates the electric flux to the charge on the plate and epsilon naught.

∫EcosθdA = q/ε_o → E2A = q/ε_o → σ = q/A →

E = σ/ε_o → E = q/2(r^2)ε_o

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Finding the Magnetic Field: Step 1

We will use Ampere-Maxwell law: ∫Bcosθdl = µ_o(I + ε_o(dΦ_B/dt))

The current in between the capacitor plates is zero, so we are really dealing with ∫Bcosθdl = µ_o ε_o(dΦ_B/dt)

The first step is to find the time derivative of magnetic flux. Let us relate the electric field to the magnetic field: dΦ_B/dt = d/dt(∫E⋅dA)

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Finding the Magnetic Field: Step 2

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Now we can substitute our equation for the time derivative of magnetic flux into the equation: ∫Bcosθdl = µ_o ε_o(dΦ_B/dt) —> ∫Bcosθdl = µ_o ε_o(IA/(r^2ε_o). Now we can find the magnetic field in between the capacitor plates

Faraday’s Law to Find the Emf of the loop

This law defines the induced emf as the time derivative of the magnetic flux. I have substituted the definition of magnetic flux as the integral of B dot dA. We can take the time derivative of the magnetic field, which does change as a function of time. In order to relate the magnetic field with time, I used Kirchhoff's Loop Rule

Substituting for Current in Magnetic Field

Now we simply substitute for current, take the derivative, and now have a function of the emf inside the wire in relation to time. (The Q/CR shouldn’t be there, whoops). Now in order to find the current, we simply divide the emf by the resistance of the wire

Final Answer

Making Sense of Our Answer

It makes sense that the current will be an exponential function, since it will become exponentially more difficult to add charge to the capacitor. Interestingly, the current doesn’t matter whatever distance you are from the plate.