B-field of ideal cylindrical solenoid, using Biot-Savart.

[SonOfOle]

Homework Statement

Consider an ideal cylindrical solenoid of length L and radius a=L/2 on which a thin wire has been wrapped a total of N turns. A steady current I flows through the wire. Assume the wires are wound so tightly that the solenoid can be thought of as a collection of N parallel current loops. Using the Biot-Savart law, find the induced magnetic field at the exact center of the solenoid.

Homework Equations

Eqn 1: [tex] B_{loop, on center axis} = \frac{\mu_{0} I (L/2)^2 }{2 ((L/2)^2 +Z^2 )^{3/2}} [/tex]

The Attempt at a Solution

First showing Eqn 1 (by using the Biot-Savart Law) we know the field due to one coil. Other than giving the solution as a sum (e.g. [tex] B_{solenoid}=\Sigma^{N}_{n=1} \frac{\mu_{0} I (L/2)^2 }{2 ((L/2)^2 +Z_{n}^{2} )^{3/2}}[/tex] where [tex] Z_{n} = L/2 - (L/N)n[/tex]), does someone see a good way to give the exact answer (we know what we're looking for from Gauss' Law, I just can't figure it out using Biot-Savarts).

 

[Nick89]

Don't sum, integrate.

Look at one loop at position z, with thickness dz.
The current through this loop is then [itex]n \, I \, dz[/itex] where n = N/L

Then, integrate from the left side to the right side.

 

[SonOfOle]

Thanks, that's helpful...

I can see from a table that [tex]\int (u^2+a^2)^{-3/2} du = u/(a^2 \sqrt{a^2 + u^2}) + C [/tex]. Any tips on how one works out that integral?

 

[SonOfOle]

nevermind. I got it. Trig substitution.

 

[SonOfOle]

The answer I got, for the record, was [tex] \mu_{0} I N /( 2 L) [/tex]. That's half of an ideal solenoid, which is reasonable given the dimensions.

 

[alphysicist]

Hi SonOfOle,

The answer I got, for the record, was [tex] \mu_{0} I N /( 2 L) [/tex]. That's half of an ideal solenoid, which is reasonable given the dimensions.

I don't think the numerical factor in that expression is correct.

 

[Nick89]

I think the answer is wrong all together, not only the numerical factor.If you put the solenoid with its center in the origin, and with its axis along the x-axis. Then you can look at one loop of current at position [itex]x[/itex] and width [itex]dx[/itex]
The current through this loop is ofcourse [tex]\frac{N \, I \, dx}{L}[/tex].

The magnetic field of a loop of current at a distance [itex]x[/itex] from its center is usually given by:
[tex]B = \frac{\mu_0 I }{2} \frac{a^2}{(x^2 + a^2)^{3/2}}[/tex]

So in this case, the magnetic field due to the one loop only is given by (replacing I with the expression for the current above):
[tex]dB_x = \frac{\mu_0 N I dx}{2L} \frac{a^2}{(x^2 + a^2)^{3/2}}[/tex]

You can now integrate this expression from -L/2 to +L/2. The answer you get should look like the usual magnetic field due to a solenoid (which you get from Ampere's law), but with an additional term. You will then see that if you make certain assumptions (that the length is much longer than the radius for example) you arrive at the expression found by Ampere's law exactly.

 

[alphysicist]

Hi Nick89,

I think the answer is wrong all together, not only the numerical factor.


If you put the solenoid with its center in the origin, and with its axis along the x-axis. Then you can look at one loop of current at position [itex]x[/itex] and width [itex]dx[/itex]
The current through this loop is ofcourse [tex]\frac{N \, I \, dx}{L}[/tex].

The magnetic field of a loop of current at a distance [itex]x[/itex] from its center is usually given by:
[tex]B = \frac{\mu_0 I }{2} \frac{a^2}{(x^2 + a^2)^{3/2}}[/tex]

So in this case, the magnetic field due to the one loop only is given by (replacing I with the expression for the current above):
[tex]dB_x = \frac{\mu_0 N I dx}{2L} \frac{a^2}{(x^2 + a^2)^{3/2}}[/tex]

You can now integrate this expression from -L/2 to +L/2.


The answer you get should look like the usual magnetic field due to a solenoid (which you get from Ampere's law), but with an additional term. You will then see that if you make certain assumptions (that the length is much longer than the radius for example) you arrive at the expression found by Ampere's law exactly.

I think you might have misread the question? They do not want the field of an infinitely long solenoid; they specifically say they want the radius to be half the length. Unless I'm reading it wrong, I believe the answer SolOfOle got was okay except the factor of 2 in the bottom should be a different number.

 

[Nick89]

Hi Nick89,

I think you might have misread the question? They do not want the field of an infinitely long solenoid; they specifically say they want the radius to be half the length. Unless I'm reading it wrong, I believe the answer SolOfOle got was okay except the factor of 2 in the bottom should be a different number.

I did not calculate the field due to an infinitely long solenoid??

Okay, I still used a as the radius, though the question states the radius is L/2, but I did that for a reason.

If you work out the integral to integrate from left to right over the length of the solenoid, you should get an answer of:
[tex]B = \mu_0 I n \frac{L}{\sqrt{L^2 + a^2}}[/tex]

Here, n is the number of turns per unit length (N/L).

Now, we know from Ampere's law that for an infinitely long solenoid, the B-field is:
[tex]B = \mu_0 I n[/tex].

In the limit that the length of the solenoid goes to infinity, we can ignore the [itex]a^2[/itex] part under the sqrt, and we are left with:
[tex]B = \mu_0 I n \frac{L}{\sqrt{L^2}} = \mu_0 I n[/tex]
Which is the same, so the expression you find from Biot-Savarts law is the same as the expression with Ampere's law, in the limit that the length goes to infinity.

This is what I was trying to say.

 

[alphysicist]

I did not calculate the field due to an infinitely long solenoid??

Okay, I still used a as the radius, though the question states the radius is L/2, but I did that for a reason.

If you work out the integral to integrate from left to right over the length of the solenoid, you should get an answer of:
[tex]B = \mu_0 I n \frac{L}{\sqrt{L^2 + a^2}}[/tex]

Here, n is the number of turns per unit length (N/L).

Now, we know from Ampere's law that for an infinitely long solenoid, the B-field is:
[tex]B = \mu_0 I n[/tex].

In the limit that the length of the solenoid goes to infinity, we can ignore the [itex]a^2[/itex] part under the sqrt, and we are left with:
[tex]B = \mu_0 I n \frac{L}{\sqrt{L^2}} = \mu_0 I n[/tex]
Which is the same, so the expression you find from Biot-Savarts law is the same as the expression with Ampere's law, in the limit that the length goes to infinity.

This is what I was trying to say.

How can you ignore the [itex]a^2[/itex] part under the square root sign, if the radius a is constrained to be half the length? As you increase the length, the radius increases, too.

 

[Nick89]

How can you ignore the [itex]a^2[/itex] part under the square root sign, if it is constrained to be half the length? As you increase the length, the radius increases, too.

I'm sorry, I see now what you mean. No, you cannot ignore the a^2 part then and you can simply plug it in. That would indeed lead to
[tex]B = c \frac{\mu_0 I N}{L}[/tex]
where c is some scalar (but not 1/2 as the OP thought).

I was merely trying to show how the expression you get from Ampere's law is the same as the expression from Biot-Savart's law in the limit that L goes to infinity, but that doesn't work here indeed, sorry!

 

[SonOfOle]

Upon further review, [tex] B = \frac{\mu_0 I N}{\sqrt{2} L} [/tex]. Thanks for your discussion Nick and AlPhysicist. (please correct me if I got my algebra wrong again).

 

[Nick89]

Upon further review, [tex] B = \frac{\mu_0 I N}{\sqrt{2} L} [/tex]. Thanks for your discussion Nick and AlPhysicist. (please correct me if I got my algebra wrong again).

I think you are correct now. :)