Magnetic field due to spiral coil

[Krushnaraj Pandya]

Homework Statement

Consider a spiral of 20 turns with inner radius R and outer radius 2R. If the current is i, find magnetic field at the center of spiral

Homework Equations

From Biot-Savart law-dB=mu(idl)/4pi*x^2

The Attempt at a Solution

Integration seems like a good option. x is continuously increasing from R to 2R, dl=xdθ. Now I'm confused as how to combine integrating dl and x...I've been thinking the whole day and turned up nothing. An interesting side question I got during the process was how to find the length of a wire used to make a spiral such as this- I feel this is somehow key to solving this problem but I'm not sure. If there is an easier method than a full blown integration, please let me know. Note- I know all the standard results such as the formula for magnetic field due to a loop, wire, arc etc. and they can be used directly if needed but I don't see anyway this can be done.
Any ideas on how to proceed with the integration or otherwise?
Also, I apologize in case I'm not able to return soon and thank you for helping me since I am buried under a huge pile of coursework- I hope you know I'm deeply grateful for the help. Thank you very much.

 

[berkeman]

Any ideas on how to proceed with the integration or otherwise?

To help you work through this conceptually, maybe try making a plot in rectangular coordinates of r(θ) where θ varies from zero to 20 * 2π...

 

[Charles Link]

If I am interpreting this one correctly, since there are 20 turns of equal thickness, the total current passing through the distance of $R$ that goes from $R$ to $2R$ is $i _{total}=20 \, i$. This means you have a current per unit length (per unit length in $r$) of $K=\frac{ 20 \, i}{R}$. $\\$ (e.g. $i_{total}=\int\limits_{R}^{2R} K \, dr$ ). $\\$ You need to apply Biot-Savart to this, but instead of $I \, dl$ in Biot-Savart, you will have $K \, r \, d \theta \, dr$ . ($K \, dr$ is the current $I$ in Biot-Savart, and $dl=r \, d \theta$). You integrate with Biot-Savart as $r$ goes from $R$ to $2R$, and $\theta$ gets integrated from $0$ to $2 \pi$.

 

[Krushnaraj Pandya]

To help you work through this conceptually, maybe try making a plot in rectangular coordinates of r(θ) where θ varies from zero to 20 * 2π...

If I am interpreting this one correctly, since there are 20 turns of equal thickness, the total current passing through the distance of $R$ that goes from $R$ to $2R$ is $i _{total}=20 \, i$. This means you have a current per unit length (per unit length in $r$) of $K=\frac{ 20 \, i}{R}$. $\\$ (e.g. $i_{total}=\int\limits_{R}^{2R} K \, dr$ ). $\\$ You need to apply Biot-Savart to this, but instead of $I \, dl$ in Biot-Savart, you will have $K \, r \, d \theta \, dr$ . ($K \, dr$ is the current $I$ in Biot-Savart, and $dl=r \, d \theta$). You integrate with Biot-Savart as $r$ goes from $R$ to $2R$, and $\theta$ gets integrated from $0$ to $2 \pi$.

I haven't encountered anything like this before.
(I'm a high school student...we are taught very basic calculus-I've managed to learn and practice more than that because of my interest in it but when dθ comes along with dr I'm totally lost.)
I understand dl=rdθ, and θ'd go from 0 to 2pi and that the r in the expression will go from r to 2r. I've heard about spherical coordinates, but I don't think its a high-school level thing. If rectangular coordinates are somewhat easier I'd like to learn though

 

[berkeman]

I understand dl=rdθ, and θ'd go from 0 to 2pi and that the r in the expression will go from r to 2r.

The problem says it takes 20 turns around the spiral to get from R to 2R, so theta goes from 0 to 20*2π (BTW, math symbols like π are under the Σ symbol in the top toolbar of the Edit window).

 

[Krushnaraj Pandya]

The problem says it takes 20 turns around the spiral to get from R to 2R, so theta goes from 0 to 20*2π (BTW, math symbols like π are under the Σ symbol in the top toolbar of the Edit window).

oh ok, that's pretty logical.
But It still won't make sense to me how to integrate two things at once, can you guide me somewhere I can learn some basic calculus like this (i.e. involving rectangular coordinates)?

 

[Charles Link]

The integration involves polar coordinates in a plane. At the origin, the magnetic field $B=\frac{\mu_o}{4 \pi} \int\limits_{R}^{2R}\int\limits_{0}^{2 \pi} \frac{K \, r \, dr \, d \theta}{r^2}$. (The OP already gave the Biot-Savart form in the original post=this is what it looks like with a current per unit length $K$). The variables of integration separate and $2 \pi$ is the result of the $d \theta$ integral. That leaves you with $B=\frac{\mu_o \, K}{2} \int\limits_{R}^{2R} \frac{dr}{r}$, with $K=\frac{20 i}{R}$. The remaining integral is somewhat elementary, and $\int\limits_{R}^{2R} \frac{dr}{r}= \ln{2}$, but yes, to understand this solution, you do need about one semester of calculus. $\\$ The current per unit length $K$ arises in geometries such as a solenoid, but otherwise, does not appear very frequently in problems. (I believe I interpreted the statement of the problem correctly, and if so, this is what is needed here in this winding which basically makes a disc shape that goes from $r=R$ to $r=2 R$). $\\$ I basically have supplied the solution here for the OP, contrary to the normal rules of the Physics Forums. Hopefully that is ok with the moderators in this instance, because the actual evaluation is somewhat routine, but it does require calculus to solve it.

 

[berkeman]

I basically have supplied the solution here for the OP, contrary to the normal rules of the Physics Forums. Hopefully that is ok with the moderators in this instance, because the actual evaluation is somewhat routine, but it does require calculus to solve it.

Yeah, it's fine in this case.

I think I would do the integration a little differently, but your method may be better.

 

[Krushnaraj Pandya]

you do need about one semester of calculus. \\

Before attempting to spend some time understanding the solution, I want to know if that's one semester of calculus beyond high school or included in it?
Also, Thank you very much for going out of your way to help me...it's something I have always loved about PF

 

[Krushnaraj Pandya]

Yeah, it's fine in this case.

I think I would do the integration a little differently, but your method may be better.

Thank you very much for your patience, I know it must be exasperating to tell me everything and me still not getting it :D
But its my view that the learning curve is directly proportional to the happiness you get when you finally learn it (at least in physics)

 

[Charles Link]

I had calculus as a senior in high school. I think it would take about 1 or two months of studying the subject for you to have enough calculus to completely understand this solution: First, you need to learn how to take derivatives. Once you have that part, integration is the opposite operation. The 2-D(double integral) integration requires just a little more effort to pick up, but that part for this problem is a simpler application.$\\$ I do think ultimately you would understand this solution a little quicker than you might expect. And once you do get enough calculus to do that, you will certainly have accomplished something.

 

[Krushnaraj Pandya]

I had calculus as a senior in high school. I think it would take about 1 or two months of studying the subject for you to have enough calculus to completely understand this solution: First, you need to learn how to take derivatives. Once you have that part, integration is the opposite operation. The 2-D(double integral) integration requires just a little more effort to pick up, but that part for this problem is a simpler application.

I have completed derivatives and integrals for trigonometric,logarithmic,polynomials,inverse trigonometric functions (and most functions I see on a day to day basis). If basic 2D integration isn't too hard, can you please give me the link to someplace I can learn more about it?

 

[Charles Link]

I have completed derivatives and integrals for trigonometric,logarithmic,polynomials,inverse trigonometric functions (and most functions I see on a day to day basis). If basic 2D integration isn't too hard, can you please give me the link to someplace I can learn more about it?

Very good. Then let me show you a couple of the steps in this problem: $\\$ $\int\limits_{R}^{2R} \int\limits_{0}^{2 \pi} \frac{r \, dr \, d \theta }{r^2}=\int\limits_{R}^{2R} \frac{dr}{r} \int\limits_{0}^{2 \pi} d \theta$. $\\$ $\int\limits_{0}^{2 \pi} d \theta=2 \pi$. $\\$ $\int\limits_{R}^{2R} \frac{dr}{r}=\ln{r}|_R^{2 R}=\ln(2R)-\ln(R)=\ln{2}$. $\\$ $K=\frac{20 \, i}{R}$ is a constant in this problem so it came outside of the integral.$\\$ I'll try to find you a "link" about double integrals. (Edit: I googled the topic: I recommend reading a calculus book for this part=the Wikipedia on the subject has too much detail). $\\$ (Additional edit: Here try this one: http://tutorial.math.lamar.edu/Classes/CalcIII/DoubleIntegrals.aspx ).$\\$ This one though is kind of simple because the variables separated with the result being a product of two integrals. When they do not separate, and when the limits are not constants, the double integrals can get a little trickier, and you might not learn that part well until you have had about 5-6 months of calculus. This one is actually one of the simpler cases.

 

[Charles Link]

@Krushnaraj Pandya $\\$ I gave this one a little more thought=it can easily be worked as a one dimensional integral $B=\frac{\mu_o}{4 \pi} \int\limits_{R}^{2R} \frac{K 2 \pi r \, dr}{r^2}$, $\\$ where the $2 \pi r$ is the circumference of a loop of width $dr$ that carries a current $K \, dr$. $\\$ In more detail, the magnetic field $dB$ from a circular loop of radius $r$ and width $dr$ is $dB=\frac{\mu_o}{4 \pi} \frac{(K \, dr \, 2 \pi r)}{r^2}=\frac{ \mu_o \, K}{2} \frac{dr}{r}$.$\\$ This just needs to get integrated from $R$ to $2R$. $\\$ The result is $B=\frac{\mu_o \, K}{2} \int\limits_{R}^{2R} \frac{dr}{r} =\frac{\mu_o K}{2} \ln{2}$.

 

[Krushnaraj Pandya]

Thank you so much,
I'm just leaving for school so I'll try to understand things when I get back but I'm grateful that you put in so much effort to help me :D
Will reply soon

 

[Charles Link]

And it could also be worked as a non-calculus problem by computing $r_j=R+(j-\frac{1}{2}) \Delta r$, where $\Delta r=\frac{R}{20}$, and $j$ is summed from $1$ to $20$. The solution is then $B=\frac{\mu_o}{4 \pi} I \sum\limits_{j=1}^{20} \frac{2 \pi r_j}{r_j^2}=\frac{\mu_o}{2} I \sum\limits_{j=1}^{20} \frac{1}{r_j}=\frac{\mu_o \, I}{2R} \sum\limits_{j=1}^{20} \frac{1}{1+(j-\frac{1}{2})(\frac{1}{20})}$.$\\$ I think if you do the last summation, you will find it gives a result that is approximately $20 \ln{2}$, so that this result would be in close agreement with our calculus result. You can do this summation very quickly with an EXCEL spreadsheet, but I currently don't have EXCEL on my computer. (Edit: I summed it by hand, making a few estimates and got around 13.7. I'd be curious to know what the more precise/exact answer of the summation is, if anyone cares to post it. Meanwhile $20 \ln{2} \approx 13.86$ ). $\\$ When you get a little more practice with the calculus, you should be able to readily show why the numerical summation gives approximately the same answer as the integral. The last summation is approximately $\int\limits_{0}^{20} \frac{dx}{1+\frac{x}{20}}=20 \int\limits_{0}^{1} \frac{ du}{1+u}= 20 \int\limits_{1}^{2} \frac{dv}{v}=20 \ln{2}$.

 

[Krushnaraj Pandya]

And it could also be worked as a non-calculus problem by computing $r_j=R+(j-\frac{1}{2}) \Delta r$, where $\Delta r=\frac{R}{20}$, and $j$ is summed from $1$ to $20$. The solution is then $B=\frac{\mu_o}{4 \pi} I \sum\limits_{j=1}^{20} \frac{2 \pi r_j}{r_j^2}=\frac{\mu_o}{2} I \sum\limits_{j=1}^{20} \frac{1}{r_j}=\frac{\mu_o \, I}{2R} \sum\limits_{j=1}^{20} \frac{1}{1+(j-\frac{1}{2})(\frac{1}{20})}$.$\\$ I think if you do the last summation, you will find it gives a result that is approximately $20 \ln{2}$, so that this result would be in close agreement with our calculus result. You can do this summation very quickly with an EXCEL spreadsheet, but I currently don't have EXCEL on my computer. (Edit: I summed it by hand, making a few estimates and got around 13.7. I'd be curious to know what the more precise/exact answer of the summation is, if anyone cares to post it. Meanwhile $20 \ln{2} \approx 13.86$ ). $\\$ When you get a little more practice with the calculus, you should be able to readily show why the numerical summation gives approximately the same answer as the integral. The last summation is approximately $\int\limits_{0}^{20} \frac{dx}{1+\frac{x}{20}}=20 \int\limits_{0}^{1} \frac{ du}{1+u}= 20 \int\limits_{1}^{2} \frac{dv}{v}=20 \ln{2}$.

Alright!
So it turns out this was easier than I was imagining it to be. I understand both the calculus solutions and they were beautiful (also you explained it really well)! These eureka moments when someone explains an elegant solution are the times when my love for physics increases :D
is the last solution a Riemann sum? I don't understand it very well, but I wish to learn more about them

 

[Charles Link]

Very good. I'm glad you enjoyed the solutions and that you were able to follow them. $\\$ In the last sum $\Delta j=1$ is the interval. The number $20$, appearing in two places, complicates things somewhat. It would be easier to look at $\int\limits_{1}^{2} \frac{dv}{v}$ and write it as a summation, than to try to explain in detail the summation connected with this slightly more complicated function. $\\$ And I'm not completely familiar with all of the calculus terms, but yes, I think that might be called a Riemann sum. (I believe that's what they call the summation that becomes an integral in the limit that $N \rightarrow +\infty$).$\\$ (Edit: I think I can readily explain this summation/integral=let's give it a try:) When the number of intervals $N$ goes to infinity, it becomes an integral, with the complicating factor in this problem being that $N$ needs to get factored out in order to make $\Delta x=\frac{1}{N}$ be the interval size, and then the summation becomes $N \sum\limits_{j=1}^{N} \frac{\Delta x}{1+x_j}$, where $x_j=(j-\frac{1}{2})(\frac{1}{N})$. $\\$ As $N$ gets large, the sum becomes the integral $\int\limits_{0}^{1} \frac{dx}{1+x}$. $\\$ The number $N=20$ is sufficiently large, that the summation, (with the $N=20$ factored out), is nearly the value of the integral ($\ln{2}$), which is the case when $N \rightarrow +\infty$ in the summation.

 

[Krushnaraj Pandya]

Very good. I'm glad you enjoyed the solutions and that you were able to follow them. $\\$ In the last sum $\Delta j=1$ is the interval. The number $20$, appearing in two places, complicates things somewhat. It would be easier to look at $\int\limits_{1}^{2} \frac{dv}{v}$ and write it as a summation, than to try to explain in detail the summation connected with this slightly more complicated function. $\\$ And I'm not completely familiar with all of the calculus terms, but yes, I think that might be called a Riemann sum. (I believe that's what they call the summation that becomes an integral in the limit that $N \rightarrow +\infty$).$\\$ (Edit: I think I can readily explain this summation/integral=let's give it a try:) When the number of intervals $N$ goes to infinity, it becomes an integral, with the complicating factor in this problem being that $N$ needs to get factored out in order to make $\Delta x=\frac{1}{N}$ be the interval size, and then the summation becomes $N \sum\limits_{j=1}^{N} \frac{\Delta x}{1+x_j}$, where $x_j=(j-\frac{1}{2})(\frac{1}{N})$. $\\$ As $N$ gets large, the sum becomes the integral $\int\limits_{0}^{1} \frac{dx}{1+x}$. $\\$ The number $N=20$ is sufficiently large, that the summation, (with the $N=20$ factored out), is nearly the value of the integral ($\ln{2}$), which is the case when $N \rightarrow +\infty$ in the summation.

I tried to understand this a few times but it seems a little above my current abilities, but Riemann sums are in my self-proclaimed coursework and sooner or later (most probably later, since I have set a large syllabus for myself to study) I will have to study them. So if there is a useful link where I can go over the basics of the intricacies between summations and calculus I'd be really grateful for it. I am a bit familiar with the procedure used in integral calculus where we divide an area into infinite rectangles each of length dx but I'm not very comfortable with it yet- as in I can't see a summation and immediately know what it'll be as an integral.
Sorry for the late reply, I wanted to make sure that I understand what you said to my level best

 

[Charles Link]

I tried to understand this a few times but it seems a little above my current abilities, but Riemann sums are in my self-proclaimed coursework and sooner or later (most probably later, since I have set a large syllabus for myself to study) I will have to study them. So if there is a useful link where I can go over the basics of the intricacies between summations and calculus I'd be really grateful for it. I am a bit familiar with the procedure used in integral calculus where we divide an area into infinite rectangles each of length dx but I'm not very comfortable with it yet- as in I can't see a summation and immediately know what it'll be as an integral.
Sorry for the late reply, I wanted to make sure that I understand what you said to my level best

The subject is at the center of integral calculus. The integral turns out to be the function whose derivative is the function that is getting summed. $\\$ Perhaps a simple and very intuitive way to see this is to compare velocity, which is $v=\frac{ds}{dt}$ and distance traveled $s$. The distance traveled is $s=\int v \, dt$. To a very good approximation, if you know the $v$ vs. $t$ function at a lot of incremental points, you could sum up the distance traveled with rectangles out of a bunch of equally spaced points on a $v$ vs. $t$ graph. The accuracy increases as you have more and more points on the graph for higher resolution.

 

[Krushnaraj Pandya]

The subject is at the center of integral calculus. The integral turns out to be the function whose derivative is the function that is getting summed. $\\$ Perhaps a simple and very intuitive way to see this is to compare velocity, which is $v=\frac{ds}{dt}$ and distance traveled $s$. The distance traveled is $s=\int v \, dt$. To a very good approximation, if you know the $v$ vs. $t$ function at a lot of incremental points, you could sum up the distance traveled with rectangles out of a bunch of equally spaced points on a $v$ vs. $t$ graph.

I can visualize that and do understand it,

with the complicating factor in this problem being that NN N needs to get factored out in order to make Δx=1NΔx=1N \Delta x=\frac{1}{N} be the interval size, and then the summation becomes NN∑j=1Δx1+xjN∑j=1NΔx1+xj N \sum\limits_{j=1}^{N} \frac{\Delta x}{1+x_j} , where xj=(j−12)(1N)xj=(j−12)(1N) x_j=(j-\frac{1}{2})(\frac{1}{N}) . \\ As NN N gets large, the sum becomes the integral 1∫0dx1+x∫01dx1+x \int\limits_{0}^{1} \frac{dx}{1+x} . \\ The number N=20N=20 N=20 is sufficiently large, that the summation, (with the N=20N=20 N=20 factored out), is nearly the value of the integral (ln2ln⁡2 \ln{2} ), which is the case when N→+∞N→+∞ N \rightarrow +\infty in the summation.

This is the part where I got lost

 

[Charles Link]

This is the part where I got lost

This one was a little tricky, and I recommend coming back to it in a couple of weeks after you have had a little more practice with integrals.$\\$ Basically on this one, if you have a current $I$ in each wire, and you increased the number of wires with current $I$ in the same space between $R$ and $2R$, you would need to raise the current density traveling in each wire, because you would need to make the wires thinner, and with less cross-sectional area in each wire. Thereby, as $N \rightarrow \infty$ for this problem, the total current in the space $R<r<2R$ becomes infinite, and the magnetic field $B \rightarrow +\infty$. $\\$ The result for the integral is $N \, \ln{2}$. As $N$ gets large, so does the integral.

 

[Krushnaraj Pandya]

This one was a little tricky, and I recommend coming back to it in a couple of weeks after you have had a little more practice with integrals.$\\$ Basically on this one, if you have a current $I$ in each wire, and you increased the number of wires with current $I$ in the same space between $R$ and $2R$, you would need to raise the current density traveling in each wire. Thereby, as $N \rightarrow \infty$ for this problem, the total current in the space $R<r<2R$ becomes infinite, and the magnetic field $B \rightarrow +\infty$. $\\$ The result for the integral is $N \, \ln{2}$. As $N$ gets large, so does the integral.

Hmm, seems best if I return to this later- at least now I know multiple ways to solve the original question using calculus, thanks a lot for that.
And I do have about 5-6 threads regarding nuclear physics questions (simple ones at that) that I'm not getting any responses to, I would be glad if you could help (you explain very well )