Answer Check: Displacement Currents and Capacitors

In summary, the problem involves a fat wire with radius a and constant current I, which forms a parallel-plate capacitor with a narrow gap of width w << a. The goal is to find the magnetic field at a distance s < a from the axis. Using equations (1), (2), and (3), the solution involves combining (1) and (2) to get J_d = I/A, and then assuming that the current in B(r) is equal to the displacement current J_d. Plugging in the values yields B(r) = (mu_0*I/A) / (2*pi*r). This solution may seem simple, but it is correct.
  • #1
TFM
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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



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

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

[tex] B(r) = \frac{\mu_0 I}{2 \pi r} [/tex] --- (3)

The Attempt at a Solution



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


[tex] J_d = \epsilon_0 \frac{1}{\epsilon_0 A}I [/tex]

I got this to cancel down into:

[tex] J_d = \frac{I}{A} [/tex]

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:

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

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

Does this look correct?

TFM
 

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  • #2
Does this look right?
 
  • #3


Your solution looks mostly correct, but there are a few things to consider. First, in your equation (2), you have used the constant current I instead of the time derivative of the current, which is what is needed for the displacement current term. So the correct equation should be:

\frac{\partial E}{\partial t} = \frac{1}{\epsilon_0 A}\frac{\partial I}{\partial t}

Next, when you cancel out the \epsilon_0 terms in (1), you should also take into account the fact that the electric field E is not constant throughout the gap, but rather varies with distance from the axis. So the correct equation for the displacement current would be:

J_d = \epsilon_0 \frac{\partial E}{\partial t} = \frac{1}{A}\frac{\partial I}{\partial t}

Then, as you correctly noted, the displacement current is equal to the current in the magnetic field equation (3), so we can write:

B(r) = \frac{\mu_0}{2 \pi r}\frac{\partial I}{\partial t}

This means that the magnetic field in the gap will vary with time, as the current is changing in time. To find the magnetic field at a specific distance s from the axis, you can use the equation:

B(s) = \frac{\mu_0}{2 \pi s}\frac{\partial I}{\partial t}

Overall, your approach is correct, but make sure to include all the necessary variables and consider the time dependence of the electric field and current.
 

Related to Answer Check: Displacement Currents and Capacitors

1. What are displacement currents?

Displacement currents are a type of electric current that is produced by changing electric fields. They are not the movement of charged particles, but rather the change in the electric field itself.

2. How are displacement currents related to capacitors?

Displacement currents play a crucial role in the functioning of capacitors. This is because capacitors store energy in the electric field between two conductors, and the changing electric field produces a displacement current.

3. Can displacement currents be detected?

Yes, displacement currents can be detected using specialized equipment such as an ammeter or a current sensor. However, they are typically very small and difficult to measure.

4. Are displacement currents the same as conduction currents?

No, displacement currents and conduction currents are two different types of electric currents. Conduction currents are the movement of charged particles, while displacement currents are caused by changing electric fields.

5. What is the importance of displacement currents in electromagnetic waves?

Displacement currents are an essential part of Maxwell's equations, which describe the behavior of electromagnetic waves. Without displacement currents, the equations would not accurately predict the behavior of electromagnetic waves.

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