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

[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?

 

[TSny]

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?

 

[peace]

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.

 

[TSny]

I don't know exactly. There must be another way.

Vertical strips

 

[peace]

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.

 

[peace]

Vertical strips

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

 

[BvU]

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

 

[peace]

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?

 

[TSny]

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?

 

[peace]

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:

 

[peace]

I turned your idea into the following shot:

But calculations with regard to this idea:

 

[TSny]

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

 

[peace]

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.

 

[peace]

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?

 

[TSny]

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]

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) ...

 

[TSny]

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]

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?

 

[TSny]

Should I put this x in the integral of Post #11?

No, use it the integral of post #14.

 

[peace]

No, use it the integral of post #14.

But I want to get integral of post #14 through integral of post #11.

 

[TSny]

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

 

[peace]

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.

 

[TSny]

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]

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 ?

 

[peace]

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.

 

[TSny]

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∅).

 

[TSny]

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]

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 .

 

[TSny]

Good

 

[peace]

Good

could I use x from the beginning?

 

[TSny]

could I use x from the beginning?

You mean the x of post #9? Yes. Try it.

 

[peace]

You mean the x of post #9? Yes. Try it.

OK. Thank you very much for your great help.

 

[TSny]

You're welcome. Enjoy your studies.