# Mutual inductance between a very long, straight wire and a Semicircular loop

• peace
In summary: Yes, exactly. Because my professor said you have to reach integral of post #14.The symbol x has a different meaning...
peace
Homework Statement
Calculate the mutual inductance between a very long, straight wire and a semicircular loop
Relevant Equations
B = µI/2πr

Φ = ∫B.dA
I think I have to assume a point like P in the semicircle. The point in terms of r and θ: P (r,θ).

So the magnetic field at that point:
B = µI/2π(R+rcosθ) .

So the magnetic flux:
Φ = ∫B.dA= µI/2π ∫∫ rdrdθ / R+rcosθ .

Is this the correct solution?

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peace said:
So the magnetic flux:
Φ = ∫B.dA= µI/2π ∫∫ rdrdθ /( R+rcosθ) .

Is this the correct solution?
I think your expression for Φ is correct. (I added parentheses in the denominator.) But the integration looks hard to me. I would avoid using polar coordinates r and θ. Can you think of another way to set up the integration over the semicircular area that takes advantage of the fact that B is independent of the vertical position inside the area?

gleem
TSny said:
I think your expression for Φ is correct. (I added parentheses in the denominator.) But the integration looks hard to me. I would avoid using polar coordinates r and θ. Can you think of another way to set up the integration over the semicircular area that takes advantage of the fact that B is independent of the vertical position inside the area?
I don't know exactly. There must be another way.

peace said:
I don't know exactly. There must be another way.
Vertical strips

TSny said:
Vertical strips
I don't think that can be done. I think the vertical strips are suitable for all shapes such as the square. But I try it now.

TSny said:
Vertical strips
What is the height of this vertical strip? The height is variable.

There is an expression for lower bound and upper bound, as a function of distance from the wire. After that
you just ##\int B\cdot dA##...

BvU said:
There is an expression for lower bound and upper bound, as a function of distance from the wire. After that
you just ##\int B\cdot dA##...
is my steps correct?

peace said:
What is the height of this vertical strip? The height is variable.

Yes, the strip's height 2y depends on x. Can you find the functional dependence of y on x?

BvU
TSny said:
View attachment 251804

Yes, the strip's height 2y depends on x. Can you find the functional dependence of y on x?
I turned your idea into the following shot:

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peace said:
I turned your idea into the following shot:
But calculations with regard to this idea:

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• Untitled.png
5.2 KB · Views: 268
Yes, looks good so far. You had a sign error in your first equation of post #10, but post #11 looks OK.

peace
TSny said:
Yes, looks good so far. You had a sign error in your first equation of post #10, but post #11 looks OK.
yes, i noticed my mistake in that equation right now. You are right.

TSny said:
Yes, looks good so far. You had a sign error in your first equation of post #10, but post #11 looks OK.

The integral should be similar to the following integral:

Φ = µI/π ∫ (√R-x / √R+x) dx

Is there a way to achieve this integral?

peace said:
The integral should be similar to the following integral:

Φ = µI/π ∫ (√R-x / √R+x) dx

Is there a way to achieve this integral?
Yes. Good. Your x here is the same as the x in my figure of post #9. Try letting x = Rcos(2θ) and recall some trig identities. Note that we're very lucky that the distance from the wire to the loop is the same as the radius of the loop! Otherwise, the integrand would not have simplified to your expression above.

peace
TSny said:
Yes. Good. Your x here is the same as the x in my figure of post #9. Try letting x = Rcos(2θ) and recall some trig identities. Note that we're very lucky that the distance from the wire to the loop is the same as the radius of the loop! Otherwise, the integrand would not have simplified to your expression above.
Ok. but the problem is how x = Rcos(2θ) ...

peace said:
Ok. but the problem is how x = Rcos(2θ) ...
Here, θ is not the polar coordinate angle. It's just a new variable of integration. I should have used a different symbol, say x = Rcos(2∅). So, you're switching the integration variable from x to ∅ by using this substitution.

peace
TSny said:
Here, θ is not the polar coordinate angle. It's just a new variable of integration. I should have used a different symbol, say x = Rcos(2∅). So, you're switching the integration variable from x to ∅ by using this substitution.
Should I put this x in the integral of Post #11?

peace said:
Should I put this x in the integral of Post #11?
No, use it the integral of post #14.

peace
TSny said:
No, use it the integral of post #14.
But I want to get integral of post #14 through integral of post #11.

Are you asking how the integral in #11 can be converted to the integral in #14?

TSny said:
Are you asking how the integral in #11 can be converted to the integral in #14?
Yes, exactly. Because my professor said you have to reach integral of post #14.

The symbol x has a different meaning in the two integrals. In post #11, you were letting x represent the distance shown here:
But, in post #14, x represents the distance shown here:

Suppose we use a capital X for your distance in post 11 and a lower case x for the distance shown for post 14. Then your integral in post 11 would look the same except for using X instead of x. To get the integral for post 14, you need to change the variable of integration from X to x. What is the relation between X and x? Use this as a substitution to get from the integral in 11 to the integral in 14. Also, be sure to include the limits for the integrals.

peace
TSny said:
The symbol x has a different meaning in the two integrals. In post #11, you were letting x represent the distance shown here:
View attachment 251815But, in post #14, x represents the distance shown here:
View attachment 251816
Suppose we use a capital X for your distance in post 11 and a lower case x for the distance shown for post 14. Then your integral in post 11 would look the same except for using X instead of x. To get the integral for post 14, you need to change the variable of integration from X to x. What is the relation between X and x? Use this as a substitution to get from the integral in 11 to the integral in 14. Also, be sure to include the limits for the integrals.
How did you find that x is the same as x in post #9 ?

X=R+x , dX=dx
so:
Φ = µI/π ∫ (√R^2 - x^2 / R+x ) dx .

But it's still different from the integral we want. But this seems to be the utmost similarity to the integral we want.

When I first worked the problem, I chose x as shown in post #9. It seemed more natural to me to do it that way. This led directly to the integral of post 14. But your choice is not wrong, it's just not as convenient. If you set up the integral as you did, then you are naturally led to making a substitution that effectively takes you to a variable of integration that represents the distance x of post #14. Then you can proceed with a further substitution x = Rcos(2∅).

peace
peace said:
X=R+x , dX=dx
so:
Φ = µI/π ∫ (√R^2 - x^2 / R+x ) dx .

But it's still different from the integral we want. But this seems to be the utmost similarity to the integral we want.
This can be simplified to what you want. Factor the expression in the square root.

peace
TSny said:
This can be simplified to what you want. Factor the expression in the square root.
Φ = µI/π ∫ (√R^2 - x^2 / R+x ) dx
Φ =µI/π ∫ √ (R-x)(R+x) / R+x ) dx
Φ =µI/π ∫ √ (R-x)√(R+x)/R+x) dx
so:
Φ =µI/π ∫ √ (R-x) / √(R+x) dx .

Good

TSny said:
Good
could I use x from the beginning?

peace said:
could I use x from the beginning?
You mean the x of post #9? Yes. Try it.

hutchphd
TSny said:
You mean the x of post #9? Yes. Try it.
OK. Thank you very much for your great help.

hutchphd

peace

## 1. What is mutual inductance?

Mutual inductance is a measure of the ability of two conductors to induce a voltage in each other due to changes in current flow. It is a fundamental concept in electromagnetism and is often used in the design of electrical circuits and devices.

## 2. How is mutual inductance between a wire and loop calculated?

The mutual inductance between a very long, straight wire and a semicircular loop can be calculated using the formula M = μ0 * N1 * N2 * (π/2), where μ0 is the permeability of free space, N1 is the number of turns in the wire, and N2 is the number of turns in the loop.

## 3. Does the distance between the wire and loop affect mutual inductance?

Yes, the distance between the wire and loop does affect mutual inductance. The closer the wire and loop are to each other, the stronger the mutual inductance will be. This is because the magnetic field from the wire will have a greater impact on the loop when they are closer together.

## 4. How does the shape of the loop affect mutual inductance?

The shape of the loop does not significantly affect mutual inductance. As long as the loop is symmetrical and the wire is very long and straight, the mutual inductance will be the same regardless of the shape of the loop.

## 5. What are some real-life applications of mutual inductance?

Mutual inductance has many practical applications, including in transformers, motors, and generators. It is also used in wireless charging technology, where mutual inductance is used to transfer energy between two devices without the need for physical contact.

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