• christof6869
In summary: This is because the current is distributed uniformly and the ratio of areas between the two wires is 4:1.Using the ampere's law equation, the magnetic field inside the larger wire is mu0(4I/3) x (r2/R2). The magnetic field inside the smaller wire is mu0(-I/3) x (r2/(R/2)^2).Combining the two fields using the superposition principle, we get B = mu0 (4I/3 x (r2/R2) - I/3 x (r2/(R/2)^2)). This simplifies to B = mu0 (I x (r2/R2) - I x (r2/(R
christof6869

Homework Statement

The conductor is cylindrically shaped with a radius of R. The hole on the inside is also cylindrically shaped with a radius of R/2. If a cross-section of this conductor is placed on an x-y coordinate plane, the hole would be centered on the x-axis at a distance of R/2 from the origin. See the attached figure. There is a current going through the conductor "out of the screen" or toward us, given as I, and it is uniformly distributed throughout the filled region. I need to find the magnetic field at any arbitrary point P inside the cavity.

Homework Equations

I know that this is an ampere's law problem similar to finding the magnetic field inside a cylindrical wire. You consider the radius of the closed loop to be less than the outer radius of the wire and, since the current is uniform, the field ends up being equivalent to the permeability of free space constant times the current enclosed multiplied by the ratio of areas.

B=mu0I x (r2/R2)

The Attempt at a Solution

I tried to go about this by imagining the wire to be a combination of two wires, where wire 1 is a larger, circular wire with the current I and the smaller circular wire has the same current in the opposite direction. By combining the two you get a hole with no current of the desired geometry. See Figure 2.

For the larger wire, a smaller radius r is used, this allows us to find the magnetic field due to the filled region of a radius r centered at the origin. I know it will be tangential to the loop I chose and that the result is going to be the same as the one in the given equations.

For the smaller wire, the loop is going to be arcing through the cross section at some arbitrary distance equivalent to r. If I convert this to polar coordinates, the area is pretty easy to find.

x=cos(theta)
y=sin(theta)
r=x2+y2

The wire can be represented by the polar equation for a circle centered on the x-axis with radius R/2.

r1 = 2(R/2)cos(theta) = Rcos(theta)

and the loop "arcing" through the circle is simply a circle, so...

r2= r

Now here's where things get sticky. In order to set up the area integral, I need to find the limits. I think I can do this by setting r1 and r2 equal to each other.

Rcos(theta) = r

cos(theta) = r/R

I want an arc who's cosine ratio is r/R and there are two angles that are equal and opposite, but I don't know how to represent them as two definite limits. This is the problem I'm having.

Once I do that integral, I'll be able to use superposition of the two tangential vectors for magnetic field lines at any arbitrary point by subtracting the magnetic field of wire 2 from wire 1.

B = B1 - B2
= mu0 (Ienclosed_1 - Ienclosed_2)
= mu0 (I x (r2/R2) - Ienclosed_2)

And that should give me what I need. Since the current is uniform throughout, I'm able to do it this way. If it wasn't I believe I would need a more general way of doing it.

Attachments

• cylindrical hole.jpg
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• Figure 2.jpg
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Last edited:
So I guess my ultimate question is how do I determine the limits and set up the integral using polar coordinates?

did you get a reply for this?

christof6869 said:

Homework Statement

The conductor is cylindrically shaped with a radius of R. The hole on the inside is also cylindrically shaped with a radius of R/2. If a cross-section of this conductor is placed on an x-y coordinate plane, the hole would be centered on the x-axis at a distance of R/2 from the origin. See the attached figure. There is a current going through the conductor "out of the screen" or toward us, given as I, and it is uniformly distributed throughout the filled region. I need to find the magnetic field at any arbitrary point P inside the cavity.

Homework Equations

I know that this is an ampere's law problem similar to finding the magnetic field inside a cylindrical wire. You consider the radius of the closed loop to be less than the outer radius of the wire and, since the current is uniform, the field ends up being equivalent to the permeability of free space constant times the current enclosed multiplied by the ratio of areas.

B=mu0I x (r2/R2)

The Attempt at a Solution

I tried to go about this by imagining the wire to be a combination of two wires, where wire 1 is a larger, circular wire with the current I and the smaller circular wire has the same current in the opposite direction. By combining the two you get a hole with no current of the desired geometry. See Figure 2.
The total current needs to be I, so the current in the larger wire should be 4I/3 and the current in the smaller wire should be -I/3.

I would suggest approaching this problem by using the Biot-Savart law instead of Ampere's law. The Biot-Savart law is more suitable for calculating the magnetic field at a point due to a current-carrying wire.

First, we need to consider the wire as a combination of two separate wires, as you have already suggested. Wire 1 is the larger wire with radius R and current I, while wire 2 is the smaller wire with radius R/2 and current -I (since it is in the opposite direction).

Next, we can use the Biot-Savart law to calculate the magnetic field at a point P inside the cavity, which is given by:

B = (μ0/4π) ∫ (Idl x ẑ)/r^2

Where μ0 is the permeability of free space, Idl is the current element along the wire, ẑ is the unit vector in the z-direction (perpendicular to the plane of the wire), and r is the distance from the current element to the point P.

Since the current is uniformly distributed throughout the filled region, we can simplify this equation to:

B = (μ0I/4π) ∫ (dl x ẑ)/r^2

Now, we can break this integral into two parts, one for each wire. For wire 1, the current element is dl = Rdθ ẑ, where θ is the angle in polar coordinates. And for wire 2, the current element is dl = (R/2)dθ ẑ. We can also express the distance r in terms of θ as r = Rcosθ.

Substituting these values into the equation, we get:

B = (μ0I/4π) ∫ (Rdθ ẑ x ẑ)/R^2cos^2θ + (R/2)dθ ẑ x ẑ/R^2cos^2θ

Simplifying this further, we get:

B = (μ0I/4π) ∫ (R^2dθ)/R^2cos^2θ + (R^2/4)dθ/R^2cos^2θ

= (μ0I/4π) ∫ dθ/cos^2θ + (μ0I/4π) ∫ (1/

1. What is a conducting wire with a cylindrical hole?

A conducting wire with a cylindrical hole is a type of wire that has a hollow center. It is commonly used in electrical circuits to conduct electricity.

2. How is a conducting wire with a cylindrical hole different from a regular wire?

A conducting wire with a cylindrical hole is different from a regular wire because it has a hollow center, which allows it to have different properties and uses. It is also typically made of a different material, such as copper or aluminum, which are good conductors of electricity.

3. What are the applications of a conducting wire with a cylindrical hole?

A conducting wire with a cylindrical hole has many applications in electrical engineering and other fields. It is commonly used in electronic devices, such as cell phones and computers, as well as in power transmission systems and electrical motors.

4. How is a conducting wire with a cylindrical hole made?

A conducting wire with a cylindrical hole is typically made by extruding a solid wire through a die, which creates the hollow center. The wire is then drawn through a series of dies to achieve the desired diameter and shape.

5. What are the advantages of using a conducting wire with a cylindrical hole?

There are several advantages to using a conducting wire with a cylindrical hole. It is typically lighter and more flexible than a solid wire, making it easier to work with. It also has a larger surface area, which can improve its conductivity and efficiency in some applications.

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