Line Integral to Verify the Magnetic Field B

[StillAnotherDave]

Homework Statement

Calculate the line integral for a magnetic field B around a square in the x-y plane.

Relevant Equations

$$\int \:B\cdot dl=\mu_0I$$

Hello folks,

I'm working on a question as follows:

I appreciate that there might be more sophisticated ways to do things, but I just want to check that my approach to the line integral is accurate. I will just give my working for the first side of the path.

So I have set up the path as a square in the x-y plane where x goes from -a to a and likewise y goes from -a to a. This leads to determining the line integral around the four sides of the square.

The first step is to give rho and phi-hat in Cartesian coordinates:

$ρ=\sqrt{x^2+y^2}$
$ϕ=−sinθi+cosθj$

This gives:

$B=\frac{\mu _0I}{2\pi }\frac{1}{\sqrt{x^2+y^2}}\left(-sin\theta i+cos\theta j\right)dx\:$

Beginning with the bottom side of the square, one would need to perform the following integral:

For the bottom side, $y =-a$ and $dl =idx$, so:

$$∫B⋅dl=-sin\theta \frac{\mu _0I}{2\pi }\int _{-a}^a\frac{1}{\sqrt{x^2+\left(-a\right)^2}}dx\:$$

Is this correct so far? If so, I should be able to complete the rest of the question. I just want to know that I haven't made a fundamental mistake in this first step.

 

[TSny]

Does $\theta$ remain constant along the path of integration?

 

[StillAnotherDave]

Does $\theta$ remain constant along the path of integration?

That's a good question. I already anticipated that if I am missing something in my working it's likely to be with working out how to integrate the $\theta$ component. But if the line integral runs in the i and j directions (thereby integrating from -a to a in each of the four paths), how does one include $\theta$ in the integral? My reading of the question is that it wants everything done in Cartesian coordinates. Does $\theta$ need to be converted or is there something else I need to do? It's my first course in vector calculus; assume the basics.

 

[TSny]

At a certain point of the path that has Cartesian coordinates (x, y), can you express $\sin \theta$ and $\cos \theta$ in terms of x and y?

Side note: You wrote $\hat \phi = -\sin \theta \hat i + \cos \theta \hat j$. The notation $\hat \phi$ indicates that the polar angle is symbolized by $\phi$. So, it would be more appropriate to write $\hat \phi = -\sin \phi\hat i + \cos \phi \hat j$. It appears that you switched the notation for the polar angle from $\phi$ to $\theta$. As long as you are clear on your choice of notation, it's ok.

 

[StillAnotherDave]

At a certain point of the path that has Cartesian coordinates (x, y), can you express $\sin \theta$ and $\cos \theta$ in terms of x and y?

Side note: You wrote $\hat \phi = -\sin \theta \hat i + \cos \theta \hat j$. The notation $\hat \phi$ indicates that the polar angle is symbolized by $\phi$. So, it would be more appropriate to write $\hat \phi = -\sin \phi\hat i + \cos \phi \hat j$. It appears that you switched the notation for the polar angle from $\phi$ to $\theta$. As long as you are clear on your choice of notation, it's ok.

So my basic understanding is that you can relate the polar angle to the Cartesian system through the following:

$x=rcos\theta$
$y=rsin\theta$

Is that the relevant connection?

P.s. thanks for highlighting the inconsistency with phi and theta.

 

[TSny]

So my basic understanding is that you can relate the polar angle to the Cartesian system through the following:

$x=rcos\theta$
$y=rsin\theta$

Is that the relevant connection?

Yes. That's what you need. Of course, you will want to express $r$ in terms of $x$ and $y$.

 

[StillAnotherDave]

Yes. That's what you need. Of course, you will want to express $r$ in terms of $x$ and $y$.

That's very helpful!

 

[StillAnotherDave]

Quick question, if a vector field contains a singularity, how does one show whether it is conservative or not? Can you still verify this by finding a scalar potential such that the gradient of the scalar potential is equivalent to the vector field?

 

[TSny]

Quick question, if a vector field contains a singularity, how does one show whether it is conservative or not? Can you still verify this by finding a scalar potential such that the gradient of the scalar potential is equivalent to the vector field?

Consider two examples.

The first is the electric field of a point charge $\mathbf E = \large \frac{kq}{r^2} \normalsize\mathbf {\hat r}$. This is a conservative field with a singularity. The line integral of $\mathbf E$ around any closed path that does not pass through the singularity is zero. The field has a potential $V = \large \frac{kq}{r}$.

The second example is the magnetic field of an infinitely long line current $\mathbf B = \large \frac{\mu_0 I}{2 \pi r} \normalsize\mathbf {\hat \theta}$. This is a non-conservative field with a singularity. The line integral of $\mathbf B$ around any closed path that encloses the current is nonzero. But the field has a potential $U = -\large \frac{\mu_0I \theta}{2 \pi}$. You can check that $\mathbf B = -\nabla U$. However, the potential is multivalued for a path that goes around the origin more than once. But I don't see why this would disqualify $U$ as a potential function.

So, it appears to me that there is not necessarily a direct relation between the existence of a scalar potential and whether or not a field with a singularity is conservative. But, I could be overlooking something. Maybe someone else can add some commentary.

 

[StillAnotherDave]

Consider two examples.

The first is the electric field of a point charge $\mathbf E = \large \frac{kq}{r^2} \normalsize\mathbf {\hat r}$. This is a conservative field with a singularity. The line integral of $\mathbf E$ around any closed path that does not pass through the singularity is zero. The field has a potential $V = \large \frac{kq}{r}$.

The second example is the magnetic field of an infinitely long line current $\mathbf B = \large \frac{\mu_0 I}{2 \pi r} \normalsize\mathbf {\hat \theta}$. This is a non-conservative field with a singularity. The line integral of $\mathbf B$ around any closed path that encloses the current is nonzero. But the field has a potential $U = -\large \frac{\mu_0I \theta}{2 \pi}$. You can check that $\mathbf B = -\nabla U$. However, the potential is multivalued for a path that goes around the origin more than once. But I don't see why this would disqualify $U$ as a potential function.

So, it appears to me that there is not necessarily a direct relation between the existence of a scalar potential and whether or not a field with a singularity is conservative. But, I could be overlooking something. Maybe someone else can add some commentary.

Interesting. Let me pose the question slightly differently to see if it helps answer what I'm looking for. If we imagine an infinitely long thin wire aligned along the z-axis. What (simple) methods could we use to determine if B is conservative outside the wire? My understanding was that if the curl of B is zero and you can form a scalar potential F such that $B=-\nabla F$ this would do the job..

 

[TSny]

If we imagine an infinitely long thin wire aligned along the z-axis. What (simple) methods could we use to determine if B is conservative outside the wire?

By the phrase "B is conservative outside the wire", do you mean that for any closed path (for which every point of the path is outside the wire), the line integral of B must equal zero? We know that for a closed path that encircles the wire, the line integral is not zero. So, B is "not conservative outside the wire".

But if we consider a restricted region of space outside the wire such that it is impossible to have a closed path inside this region that encloses the wire, then the line integral of B for any closed path in this region would be zero. So, you could say that B is conservative inside this region.

My understanding was that if the curl of B is zero and you can form a scalar potential F such that $B=-\nabla F$ this would do the job.

For the wire, the curl of B is zero at every point outside the wire. We also know that we can define a function F such that $B=-\nabla F$. Yet, B is not conservative outside the wire. But F is not single-valued in this case.

I think that the following is true: If we have a region of space for which the curl of B is zero at every point of the region and if you can form a single-valued potential F such that $B=-\nabla F$ at each point in the region, then B is conservative in that region. That is, the line integral of B around any closed path inside the region will be zero.