# Homework Help: Going from cylindrical to cartesian coordinates

1. Oct 19, 2012

### Niles

1. The problem statement, all variables and given/known data
Hi

The expression for the magnetic field from an infinite wire is
$$\boldsymbol B(r) = \frac{\mu_0I}{2\pi}\frac{1}{r} \hat\phi$$
which points along $\phi$. I am trying to convert this into cartesian coordinates, and what I get is
$$\boldsymbol B(x, y) = \frac{\mu_0I}{2\pi}\frac{1}{\sqrt{x^2+y^2}} \hat\phi$$
where
$$\hat\phi = -\sin\phi \hat x + \cos\phi \hat y$$
I am trying to make a phase plot of this expression, so what I have done is to say that $\phi = \arctan(y/x)$, so $\hat\phi = -\sin(\arctan(y/x))\hat x + \cos(\arctan(y/x))\hat y$. However I don't get the desired result. Have I missed something in my approach?

2. Oct 19, 2012

### voko

sin (arctan a) and cos (arctan a) can be simplified. Let z = arctan a, that means tan z = a, and tan z = sin z/cos z = a, so you can express sin z and cos z in terms of a.

3. Oct 19, 2012

### Niles

Thanks, so I know that
$$\frac{y}{x} = \frac{\sin(\phi)}{\cos(\phi)}$$
I can't see how this enables me to rewrite e.g. $\sin(\arctan(y/x))$.

4. Oct 19, 2012

### voko

$\sin (\arctan a) = \sin z$. Since $\tan z = \sin z/\cos z = a$, $\sin^2 z = a^2 \cos^2 z = a^2(1 - \sin^2 z)$. So you can find $\sin z$ as a function of $a$; ditto for $\cos z$. Then substitute $a = y/x$.

5. Oct 19, 2012

### Niles

Ah, I see. So I get
$$\sin z = \frac{a}{\sqrt{1+a^2}} \\ \cos z = \frac{1}{\sqrt{1+a^2}}$$
But I still have my original problem: That when I plot B using these for negative x, then I don't see the correct magnetic field. I thought that I was perhaps missing a term $\pi/2$, but that didn't solve it either.

6. Oct 19, 2012

### voko

What do you get and what is your expectation?

7. Oct 19, 2012

### Niles

I have attached a plot of what I see (the axes are (x, y), the current 1A and the units on the axis in meters), it is called "negative_x". If I only plot for positive x-values I get "positive_x", and there I see what I expect (as shown here, on the top: http://www.netdenizen.com/emagnet/solenoids/frommaxwellonly.htm).

My code in Mathematica for plotting is:

VectorPlot[(mu0/2 pi)*
current*(1/(x^2 + y^2)^(1/2))*{-(y/x)/(1 + y^2/x^2)^(1/2),
1/(1 + y^2/x^2)^(1/2)}, {x, -0.01, 0.01}, {y, -0.01, 0.01}]

#### Attached Files:

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• ###### negative_x.jpeg
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8. Oct 19, 2012

### voko

You should be able to simply the formula very significantly. Note that $\frac 1 {\sqrt {1 + y^2/x^2}} = \frac x {\sqrt {x^2 + y^2}}$, and the radical in the denominator nicely couples with that in the common factor. But even then your formula is not wrong, I am not sure why Mathematica does not plot it correctly.

9. Oct 19, 2012

### Niles

I don't know either. Strange, but nice to know that I have the correct exprssion. Thanks!

10. Oct 19, 2012

### Niles

OK, I just plotted it in MatLAB, and it *isn't* correct. For x<0 the y-coordinates all have to change sign. So the expression is not correct.

EDIT: I have attached the plot.

#### Attached Files:

• ###### untitled.jpg
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12.8 KB
Views:
102
Last edited: Oct 19, 2012
11. Oct 19, 2012

### voko

Have you tried the simplified formula as I suggested?

12. Oct 19, 2012

### Niles

Yes, it didn't change anything. It shouldn't either, since it is just a different way of expressing it.

13. Oct 20, 2012

### voko

Your formula is $(\mu_0/2 \pi) I \frac 1 {(x^2 + y^2)^{1/2}} \left(\frac {-y/x } {(1 + y^2/x^2)^{1/2}}, \frac 1 {(1 + y^2/x^2)^{1/2}} \right)$

Observe that the y-component is always positive, which is incorrect. If you transform it the way I suggested, you will get $(\mu_0/2 \pi) I \frac 1 {x^2 + y^2} \left(-y, x\right)$, which restores the correct sign.

14. Oct 21, 2012

### Niles

thanks! I must have made an error somewhere then when I tried