How Does Faraday's Law Apply to a Non-Uniformly Charged Cylindrical Shell?

In summary, we are discussing an infinitely long cylindrical shell with inner radius A, outer radius B, and carrying a non-uniform current density J=[2ar]sin(wt), where a and w are constants. We are asked to find the magnitude of the induced electric field at a distance r>B from the central axis. To solve this, we first integrate J with respect to r from A to B to find the total current, then use Ampere's Law to find the magnetic field. We then multiply this by an area, take the time derivative, and divide by the curve where the electric field is, due to symmetry. The induced emf is calculated using Faraday's Law, considering a tiny circle perpendicular to the field lines
  • #1
1stepatatime
12
0
First off, I'm sorry but I am LaTeX illiterate :frown:.

Homework Statement


An infinitely long conducting cylindrical shell has an inner radius A, an outer radius B, and carries a non-uniform current density J= [2ar]sin(wt) where a and w are constant.

What is the magnitude of the induced electric field a distance r > B from the central axis of the cylindrical shell.

Homework Equations



Ampere's Law
[itex] \oint {B \cdot ds = \mu _0 I_C } [/itex]

Faraday's Law
[itex] \oint {E \cdot ds = - \frac{d}{{dt}}} \Phi_B [/itex]I know the electric and magnetic field as well as the curves are all vectors, but don't know how to throw them into latex.

The Attempt at a Solution



I had a similar question like this on a final exam, and have to admit I was confused. I just don't want to bother my professor during Winter break with this question out of respect.

I think I know how to solve this mechanically, first I'd integrate the current density J with respect to r from radius A to radius B to solve for the total current making sure I include (2*pi*rdr) as a differential, then I'd throw this result into Ampere's Law to find the magnitude of the magnetic field. Then I'd have to multiple this by an area, take the time derivative, then divide out by the curve where the electric field is. This is all due to everything being symmetric.

My question is, since the field lines have to pass through a closed loop in order for there to be a flux through an area which produces an induced emf around that closed loop according to Faraday's Law. For this example isn't that area surrounded by the closed loop and the magnetic field from Ampere's Law in the same plane? How can there be flux if the field lines are parallel to the plane?

In other words, to find the flux I'd have to multiply the magnetic field by pi*r^2?. Is this, my original assumption, or neither correct?
 
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  • #2
1stepatatime said:
My question is, since the field lines have to pass through a closed loop in order for there to be a flux through an area which produces an induced emf around that closed loop according to Faraday's Law. For this example isn't that area surrounded by the closed loop and the magnetic field from Ampere's Law in the same plane? How can there be flux if the field lines are parallel to the plane?

You need to take a closed loop that is perpendicular to the field lines. Consider a tiny circle centered around "r", that's perpendicular to the magnetic field lines. It has area pi*R^2 and the induced electric field around the loop is constant (because the circle is tiny), so you can use Faraday's law of induction.
 
  • #3
ideasrule said:
You need to take a closed loop that is perpendicular to the field lines. Consider a tiny circle centered around "r", that's perpendicular to the magnetic field lines. It has area pi*R^2 and the induced electric field around the loop is constant (because the circle is tiny), so you can use Faraday's law of induction.

ahhhh I see, it makes clear sense! I wasn't thinking about "r" being in a different plane and limited it to the plane with the cross-sectional area of the cylindrical shell. Thank you Ideasrule. In regards to the flux, would you multiply the magnetic field by pi*B^2 which is the radius of the shell, from it's center even though it's a shell with an inner radius A?
 
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Related to How Does Faraday's Law Apply to a Non-Uniformly Charged Cylindrical Shell?

1. What is Faraday's Law?

Faraday's Law is a fundamental law of electromagnetism that describes the relationship between a changing magnetic field and an induced electric field.

2. How does Faraday's Law work?

Faraday's Law states that when there is a changing magnetic field through a conductor, an electric field is induced in the conductor. This electric field can then cause a current to flow if the conductor is part of a closed circuit.

3. What is the importance of Faraday's Law?

Faraday's Law is important because it explains how generators and motors work, and it is also the basis for many other important laws and principles in electromagnetism.

4. What are some real-world applications of Faraday's Law?

Some real-world applications of Faraday's Law include generators, transformers, and induction cooktops. It is also used in many other technologies, such as electric motors, power plants, and wireless charging systems.

5. How does Faraday's Law relate to Lenz's Law?

Faraday's Law and Lenz's Law are closely related and often referred to together. Lenz's Law is a consequence of Faraday's Law and states that the direction of the induced current is always such that it opposes the change that produced it.

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