How Is Flux Calculated in a Toroid with Variable Radius?

[Knissp]

Homework Statement

The question gives a coil of N turns carrying a current of I Amperes wound on a ring with rectangular cross section of inner radius r1 and outer radius r2 and height h. The ring has magnetic permeability mu. What is the flux in webbers?

Homework Equations

Ampere's Law:
Closed Integral (B * ds) = mu * I

The Attempt at a Solution

My problem is mostly a conceptual one since I have all the equations for sure.

flux = closed integral (B * dA)
integral (B * ds) = B integral (ds) = B (2 pi r) = mu * N * I
since it has N loops

B is not constant, so for flux we must integrate:
flux = integral (B dA) = integral (mu*N*I/(2 pi r) dA)

This is where I'm confused:
the solution says:
"The area element dA can be expressed as dA = h dr where h is the height of the rectangular cross section."
Then it proceeds to integrate from r1 to r2 to get: flux = N*I*h/(2 pi)*mu*ln(r2/r1) and the question is solved. But I don't understand how that substitution can be made. Doesn't that imply dA/dr = h so A is a function of r, or A = hr + c??

Any help will be appreciated greatly.

 

[anathema]

First of all, it's great that you have all the equations and are working towards a solution! Let's go through the problem step by step to understand the solution better.

1. First, let's define the problem. We have a coil of N turns carrying a current I, wound on a ring with a rectangular cross section. The ring has inner radius r1, outer radius r2, and height h. We want to find the flux in webbers.

2. Next, let's identify the relevant equation for this problem. From Ampere's Law, we know that the closed integral of B * ds is equal to mu * I. This means that the magnetic field B is related to the current I and the magnetic permeability mu.

3. Now, let's think about the flux. Flux is defined as the closed integral of B * dA, where dA is the area element. In this case, the area element is a small rectangle on the surface of the ring. Since the ring has a rectangular cross section, the area element can be expressed as dA = h * dr, where h is the height of the rectangular cross section and dr is a small change in radius.

4. We can now substitute this expression for dA into the equation for flux: flux = integral (B * dA) = integral (B * h * dr). Since we know that B is related to I and mu, we can substitute in the expression from Ampere's Law: flux = integral (mu * I * h * dr).

5. To solve this integral, we need to determine the limits of integration. We know that the magnetic field B is not constant, so we need to integrate over the entire surface of the ring. This means that we need to integrate from the inner radius r1 to the outer radius r2.

6. Now, we can solve the integral: flux = integral (mu * I * h * dr) = mu * I * h * integral (dr) = mu * I * h * (r2 - r1).

7. Finally, we can simplify this expression to get the final solution: flux = mu * N * I * h * (r2 - r1) / (2 * pi).

I hope this helps to clarify the solution for you. The key concept here is that the area element dA is a small rectangle on the surface of the ring, and can be expressed as dA

 

[Moetasim]

It is correct that the area element dA can be expressed as dA = h dr, where h is the height of the rectangular cross section. This is because in a toroid, the magnetic field lines are confined to the hollow cylindrical region, and the flux through this region is given by the integral of B*dA. Since the area of the cross section is proportional to the radius (A = h*r), we can express dA as h dr.

In this case, we are integrating over the entire cross section from r1 to r2, which means that the limits of integration for r are from r1 to r2. This does not imply that A is a function of r, but rather that the area element dA changes with r as we integrate over the different radii.

To clarify, the expression dA/dr = h does not mean that A itself is a function of r. It simply means that as we change the radius, the area element dA also changes proportionally, with a constant factor of h. Therefore, we can use this substitution in the integral to simplify the calculation.

I hope this helps to clarify the concept. Let me know if you have any further questions.