Answer Check: Displacement Currents and Capacitors

AI Thread Summary
The discussion focuses on calculating the magnetic field in a gap of a wire that acts as a parallel-plate capacitor. The user combines equations for displacement current and electric field to derive a relationship, ultimately equating the magnetic field to the displacement current. They simplify the expression to find B(r) in terms of the current I and area A. The user seeks validation for their approach and results, questioning if their assumptions and calculations are correct. The thread emphasizes the application of electromagnetic theory to solve the problem.
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Homework Statement



A fat wire, radius a, carries a constant current I, uniformly distributed over its cross section. A narrow gap in the wire, of width w << a, forms a parallel-plate capacitor, as shown in the figure above. Find the magnetic field in the gap, at a distance s < a from the axis.

{Figure given below}

Homework Equations



displacement current, J_d = \epsilon_0 \frac{\partial E}{\partial t} --- (1)

\frac{\partial E}{\partial t} = \frac{1}{\epsilon_0 A}I --- (2)

B(r) = \frac{\mu_0 I}{2 \pi r} --- (3)

The Attempt at a Solution



Okay so firstly, I have put together (1) and (2) to get:


J_d = \epsilon_0 \frac{1}{\epsilon_0 A}I

I got this to cancel down into:

J_d = \frac{I}{A}

I then made a very bold assumption that the current in B(r) = the displacement current J_d

So I Inserted the values and got:

B(r) = \frac{\mu_0 \left( \frac{I}{A}\right)}{2 \pi r}

This seems very quick and straight forwards, though...

Does this look correct?

TFM
 

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