# Thin heartshaped wire - Calculate the magnetic field at the origin

• JoelKTH
Thanks for your reply @Steve4Physics @berkeman and @haruspex . I rewrote it now. I hope its more clear.

#### JoelKTH

Homework Statement
A thin wire that carries the current I is bent in a heart-shaped curve according to the equation r_c=r_0 e^k|phi| for |phi|<= pi, where r_c is the distance from origo to a point on the curve. Calculate the magnetic flow B in origo.
Relevant Equations
Biot Savarts Law
Hi,

So I know I am to use Biot Savarts law dB= (my_0/4pi)* (I dl x (r-r')/|r-r'|^3 where r=0 because its in origo and r'=r'_c(r'_hat).
This makes (r-r')= -r'_c(r'_hat) and |r-r'|^3= r_c^3.

From previous questions, I have defined dl' as the infinitesimal displacement of r'(phi) when phi' is increasing with dphi along the curve.
Usually its dl'/dphi = r_hat --> dl= r_hat dphi and then I use Biot Savarts law. However I get wrong result here.
My professor tells me to use dl=r_c' dphi' phi_hat + dr'_c r_hat_c.
Why is this intuitively correct? I am having some problem wrapping my head around this. How can I derive it from the dl' expression? Or know if there are any other shapes that do not only depend on phi.

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JoelKTH said:
Homework Statement:: A thin wire that carries the current I is bent in a heart-shaped curve according to the equation r_c=r_0 e^k|phi| for |phi|<= pi, where r_c is the distance from origo to a point on the curve. Calculate the magnetic flow B in origo.
Relevant Equations:: Biot Savarts Law

Biot Savarts law dB= (my_0/4pi)* (I dl x (r-r')/|r-r'|^3
I think you have mismatched parens in that last equation in the quote, but it's hard to tell without LaTeX.

It would help a lot if you could start posting your math equations using LaTeX -- please see the "LaTeX Guide" at the lower left of the Edit window.

Also, I'm having trouble following where you are taking the cross product of ## \vec I(\phi)## and ## \vec r(\phi) ## -- am I missing that somewhere in your text equations? The angle of the vector cross-product looks to be a pretty complicated function of ##\phi## ...

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• MatinSAR
• nasu
Hi @JoelKTH. I'd like to add to what @berkeman and @haruspex have said.

Your use of primed and unprimed variables is confusing. For example, it is not clear what you mean by dl and dl’. I suspect there is only one line element length to consider, not two. A diagram showing the various vectors and angles might help.

JoelKTH said:
My professor tells me to use dl=r_c' dphi' phi_hat + dr'_c r_hat_c.
Why is this intuitively correct?
It looks like your professor is telling you to treat the ( vector) line element (dl) as the sum of 2 components: one component in the 'phi-hat' direction and the other component is in the 'r_hat_c' direction. But without a proper diagram showing the vectors and angles, it's hard to know if that's correct.

• MatinSAR
Thanks for your reply @Steve4Physics @berkeman and @haruspex . I rewrote it now. I hope its more clear.
JoelKTH said:
Homework Statement:: A thin wire that carries the current I is bent in a heart-shaped curve according to the equation r_c=r_0 e^k|\varphi| for |\varphi|\le \pi, where r_c is the distance from origo to a point on the curve. Calculate the magnetic flow B in origo.
Relevant Equations:: Biot Savarts Law

Hi,

So I know I am to useBiot Savart's law $$\mathbf{d B} = \mu_0 \frac{4 \pi I \mathbf{dl \times (r - r')}}{|\mathbf{r - r'}|^3}$$where $$\mathbf{r} = \mathbf{0}$$ because it is in the origin and $$\mathbf{r'} = r'_c \hat{\mathbf{r'}}$$ This makes $$(\mathbf{r - r'}) = -r'_c \hat{\mathbf{r'}}$$ and $$|\mathbf{r - r'}|^3 = r_c^3$$.

JoelKTH said:
From previous questions, I have defined $$\mathbf{dl'}$$ as the infinitesimal displacement of $$\mathbf{r'(\varphi)}$$ when $$\varphi'$$ is increasing with $$d\varphi$$ along the curve. $$\varphi : 0 \rightarrow 2\pi$$ Usually, it is $$\frac{\mathbf{dl'}}{d\varphi} = \hat{\mathbf{r}} \Rightarrow \mathbf{dl'} = \hat{\mathbf{r}} d\varphi$$ and then I use Biot Savart's law. However, I get the wrong result here.

JoelKTH said:
My professor tells me to use $$\mathbf{dl'} = r_c' \hat{\mathbf{\varphi'}} d\varphi' + dr'_c \hat{\mathbf{r'_c}}$$. Why is this intuitively correct? I am having some difficulty understanding this. How can I derive it from the $$\mathbf{dl'}$$ expression? Or are there any other shapes that do not only depend on $$\varphi$$?

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• berkeman
Steve4Physics said:
Hi @JoelKTH. I'd like to add to what @berkeman and @haruspex have said.

Your use of primed and unprimed variables is confusing. For example, it is not clear what you mean by dl and dl’. I suspect there is only one line element length to consider, not two. A diagram showing the various vectors and angles might help.
It's only dl’. Not sure what you mean by a diagram of showing the various vectors and angles. It might be a bit confusing as @haruspex also let me know. Hopefully the edited version is more straight forward.
Steve4Physics said:
It looks like your professor is telling you to treat the ( vector) line element (dl) as the sum of 2 components: one component in the 'phi-hat' direction and the other component is in the 'r_hat_c' direction. But without a proper diagram showing the vectors and angles, it's hard to know if that's correct.
Yes, I think due to the symmetry of the wire

I would like to add that there is a tip in the solution to go to kartesian coordinates.
$$\frac{d \mathbf{r'_c}}{d\varphi'} = \frac{d}{d\varphi'} (\mathbf{\hat{x}} \cos{\varphi'} + \mathbf{\hat{y}} \sin{\varphi'}) = -\mathbf{\hat{x}} \sin{\varphi'} + \mathbf{\hat{y}} \cos{\varphi'} = \mathbf{\hat{\varphi'}}$$
$$d\mathbf{l'} = \left(\mathbf{\hat{r_c}} \frac{d \mathbf{r'_c}}{d\varphi'} + \mathbf{r'_c} \frac{d \mathbf{\hat{r_c}}}{d\varphi'}\right)d\varphi'$$

Does this make more sense than using $$\mathbf{dl'} = r_c' \hat{\mathbf{\varphi'}} d\varphi' + dr'_c \hat{\mathbf{r'_c}}$$
?

I assume you were trying to fix the LaTeX. You need to bracket it with double hash signs (#, but two together) or double $signs. Also, no boldface or italics inside the LaTeX except by using LaTeX controls. As I wrote in post #3, I have no idea where you get ##\vec {dl}=\hat r d\phi## from. That makes no sense to me. As @Steve4Physics noted in post #4, your professor's version looks to be based on writing the length element vector as the sum of a radial component and a tangential component, ##\vec{dl}=dr.\hat r+r.d\phi.\hat \phi##, which is clearly correct. • MatinSAR @JoelKTH, in Post #5 you wrote: ##\mathbf{d B} = \mu_0 \frac{4 \pi I \mathbf{dl \times (r - r')}}{|\mathbf{r - r'}|^3}## but the ##4\pi## should be in the denominator. Note that to find the field at the origin you can use simpler equations: ##\mathbf {d B} = \frac {\mu_0 I}{4 \pi} \frac {\mathbf{dl} \times \mathbf r}{|\mathbf{r^3}|}## where ##r = r_0 e^{k|\phi|}##, ##-\pi \le \phi \lt \pi##. Then you don't need to use variables which are primed or subscripted with ‘c’ - which reduces headaches. • JoelKTH and MatinSAR haruspex said: I assume you were trying to fix the LaTeX. You need to bracket it with double hash signs (#, but two together) or double$ signs. Also, no boldface or italics inside the LaTeX except by using LaTeX controls.

As I wrote in post #3, I have no idea where you get ##\vec {dl}=\hat r d\phi## from. That makes no sense to me. As @Steve4Physics noted in post #4, your professor's version looks to be based on writing the length element vector as the sum of a radial component and a tangential component, ##\vec{dl}=dr.\hat r+r.d\phi.\hat \phi##, which is clearly correct.
Thanks for your reply. Yes the ##\vec {dl}=\hat r d\phi## is wrong. Instead it normally should be: ##\vec {dl}=\hat \varphi d\varphi##. Due to cylindrical symmetry(no heart shape) ##\vec {dl}=\hat r d\phi##.

But now its heart-shaped. I suppose it makes more sense that if the radius is not constant, an infinitesimal movement around the wire should then also be dependent on the radius. Which most likely is the ##r.d\phi.\hat \phi## part. I think this is where I am confused and perhaps its more a question of vector calculus than physics.

After studying my Mathematics Handbook of 600 pages that contains vector component relationship, I see the the patter. Thanks for your help

JoelKTH said:
it normally should be: ##\vec {dl}=\hat \varphi d\varphi##.
No, that can't be right either; it's dimensionally wrong. You have a length on the left but not on the right.
For constant radius it would be: ##\vec {dl}=r\hat \varphi d\varphi##.
Since the radius changes too, there is also the vector due to the radius change: ##\hat r.dr##.
Hence, in general, ##\vec {dl}=\hat \varphi r.d\varphi+\hat rdr##.
If ##r=r_0e^{k\varphi}##, can you express ##dr## in terms of ##d\varphi##?

haruspex said:
No, that can't be right either; it's dimensionally wrong. You have a length on the left but not on the right.
For constant radius it would be: ##\vec {dl}=r\hat \varphi d\varphi##.
Since the radius changes too, there is also the vector due to the radius change: ##\hat r.dr##.
Hence, in general, ##\vec {dl}=\hat \varphi r.d\varphi+\hat rdr##.
If ##r=r_0e^{k\varphi}##, can you express ##dr## in terms of ##d\varphi##?
Yes, you are correct. I ment what you wrote but I was a bit too quick here in Latex.
Not sure what you mean by expressing ##dr## in terms of ##d\varphi##?

I have ##\vec {dl}=\hat \varphi r.d\varphi+\hat rdr##. I know that $$d\mathbf{l'} = \left(\mathbf{\hat{r_c}} \frac{d \mathbf{r'_c}}{d\varphi'} + \mathbf{r'_c} \frac{d \mathbf{\hat{r_c}}}{d\varphi'}\right)d\varphi'$$
Therefor, I derive each respective part. The first part I go to cartesian coordinates to simplify to ##\varphi## direction

JoelKTH said:
Not sure what you mean by expressing ##dr## in terms of ##d\varphi##?

Differentiate ##r=r_0e^{k\varphi}##.

I have an issue with the function ##r=r_0 e^{k|\phi |} ## representing a closed electrical circuit. As the angle increases the value of ##r## increases exponentially and doesn't close a circuit. The picture you drew appears to be a cardioid whose equation is,
$$r=2a(1-\cos(\phi))$$

Fred Wright said:
I have an issue with the function ##r=r_0 e^{k|\phi |} ## representing a closed electrical circuit. As the angle increases the value of ##r## increases exponentially and doesn't close a circuit.
Try ##-\pi<\phi<\pi##, as stated in post #1.

haruspex said:
Differentiate ##r=r_0e^{k\varphi}##.
##r=r_0ke^{k\varphi}##

JoelKTH said:
##r=r_0ke^{k\varphi}##
No, that's incorrect differentiation. You have to pick what it is you are differentiating with respect to and do it consistently both sides.
If I differentiate wrt phi: ##\frac{dr}{d\phi}=r_0ke^{k\varphi}##, which I can rewrite as ##dr=r_0ke^{k\varphi}d\phi##.
You are now in a position to find ##\vec dl## in terms of ##\phi, d\phi## and the unit vectors.

haruspex said:
No, that's incorrect differentiation. You have to pick what it is you are differentiating with respect to and do it consistently both sides.
If I differentiate wrt phi: ##\frac{dr}{d\phi}=r_0ke^{k\varphi}##, which I can rewrite as ##dr=r_0ke^{k\varphi}d\phi##.
You are now in a position to find ##\vec dl## in terms of ##\phi, d\phi## and the unit vectors.
Yes you are correct. Here is my calculation:

##d\mathbf{l}' = d\mathbf{r} = (r_c)\hat{\mathbf{d}}r_c' + (\varphi)\hat{\mathbf{r}}_c' d\varphi'##

Then, ##dr/d\varphi = \hat{\mathbf{r}}_c \frac{dr_c'}{d\varphi} + \hat{\boldsymbol{\varphi}} \frac{r_c' d\varphi'}{d\varphi'}##

Since ##r_c = r_0 e^{k|\varphi|}##, we have ##\frac{d r_c'}{d\varphi'} = r_0 k e^{k|\varphi|} = r_c k##

By transferring to cartesian coordinates we get ##\frac{dr_c'}{d\varphi} = \frac{d}{d\varphi} ( \hat{\mathbf{x}} \cos\varphi' + \hat{\mathbf{y}} \sin\varphi') = \hat{\boldsymbol{\varphi}}##

This means that ##\mathrm{d}\mathbf{l}'=(r_c\hat{\mathbf{d}}r_c'/d\varphi+\varphi\hat{\mathbf{r}}_c'/d\varphi')d\varphi'=(r_c\hat{\mathbf{r}}_ckr_c'+r_c'\varphi\hat{\boldsymbol{\varphi}})##

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JoelKTH said:
Yes you are correct. Here is my calculation:

##d\mathbf{l}' = d\mathbf{r} = (r_c)\hat{\mathbf{d}}r_c' + (\varphi)\hat{\mathbf{r}}_c' d\varphi'##

Then, ##dr/d\varphi = \hat{\mathbf{r}}_c \frac{dr_c'}{d\varphi} + \hat{\boldsymbol{\varphi}} \frac{r_c' d\varphi'}{d\varphi'}##

Since ##r_c = r_0 e^{k|\varphi|}##, we have ##\frac{d r_c'}{d\varphi'} = r_0 k e^{k|\varphi|} = r_c k##

By transferring to cartesian coordinates we get ##\frac{dr_c'}{d\varphi} = \frac{d}{d\varphi} ( \hat{\mathbf{x}} \cos\varphi' + \hat{\mathbf{y}} \sin\varphi') = \hat{\boldsymbol{\varphi}}##

This means that ##\mathrm{d}\mathbf{l}'=(r_c\hat{\mathbf{d}}r_c'/d\varphi+\varphi\hat{\mathbf{r}}_c'/d\varphi')d\varphi'=(r_c\hat{\mathbf{r}}_ckr_c'+r_c'\varphi\hat{\boldsymbol{\varphi}})##
Why do you insist on using all these primes (‘) in your notation? Don't you get fed up typing them? I don't see what the c subscript gains either.
To avoid blunders, we should consider ##\phi>0## and ##\phi<0## separately.
We have for positive ##\phi##:
##r=r_0e^{k\phi}##
##dr=kr_0e^{k\phi}d\phi##
##d\vec l= \hat\phi r.d\phi+\hat rdr##
Can you combine those to eliminate r, i.e. to obtain ##d\vec l## as a function of the two unit vectors and ##r_0, k, \phi, d\phi## only?
Similarly, write ##\vec r## in those terms and compute ##d\vec l\times\vec r##.