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Fernando Garcia

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PHYS405 - Electricity & Magnetism I

Spring 2024

Problem 1.65

Let's recall The Fundamental Theorem for curls, also known as Stokes' theorem:

$$\begin{array}{}\text{(Eq. 1.57, page 33)}& {\int }_{\mathcal{S}}(\mathrm{\nabla }\times \mathbf{v})\cdot d\mathbf{a}={\oint }_{\mathcal{P}}\mathbf{v}\cdot d\mathbf{l}\end{array}$$

(a) Let's consider the circular path (of radius $R$, centered at the origin)

$$\mathbf{r}(t)=\left(\begin{array}{c}R\cos \left(t\right)\\ R\sin \left(t\right)\end{array}\right)$$

Where $t$ will go from $0$ to $2\pi$. The field $\mathbf{A}$ at some point on the curve given by $t$ will be given by:

$$\begin{array}{rl}\mathbf{A}& =-\frac{R\sin \left(t\right)}{R^2{\cos }^{2}t+R^2{\sin }^{2}t}\hat{\mathbf{x}}+\frac{R\cos \left(t\right)}{R^2{\cos }^{2}t+R^2{\sin }^{2}t}\hat{\mathbf{y}}\\ & =-\frac{R\sin \left(t\right)}{R^2}\hat{\mathbf{x}}+\frac{R\cos \left(t\right)}{R^2}\hat{\mathbf{y}}\\ & =-\frac{\sin \left(t\right)}{R}\hat{\mathbf{x}}+\frac{\cos \left(t\right)}{R}\hat{\mathbf{y}}\end{array}$$

To compute the line integral, let's first compute the differential $d\mathbf{r}$:

$$d\mathbf{r}=\left(\begin{array}{c}-R\sin \left(t\right)\\ R\cos \left(t\right)\end{array}\right)dt$$

We can now compute the line integral:

$$\begin{array}{rl}{\oint }_{\mathcal{C}}\mathbf{A}\cdot d\mathbf{r}& ={\int }_{t=0}^{2\pi }\left(\begin{array}{c}-\sin \left(t\right)/R\\ \cos \left(t\right)/R\end{array}\right)\cdot \left(\begin{array}{c}-R\sin \left(t\right)\\ R\cos \left(t\right)\end{array}\right)dt\\ & ={\int }_{t=0}^{2\pi }{\sin }^{2}t+{\cos }^{2}tdt\\ & ={\int }_{t=0}^{2\pi }1dt\\ & =2\pi \end{array}$$

Now, let's consider the surface integral of the curl. It is straightforward to see that the curl of $\mathbf{A}$ is

$$\mathrm{\nabla }\times \mathbf{A}=\mathbf{0}$$

So... the surface integral is 0! This contradicts Stokes' theorem (since the line integral yielded $2\pi$, not 0).

To fix this, let's rewrite the curl using Dirac Deltas:

$$\mathrm{\nabla }\times \mathbf{A}=2\pi \delta (x)\delta (y)\hat{\mathbf{z}}$$

When doing so, we see that

$$\begin{array}{rl}{\int }_{\text{Circle of radius }R}(\mathrm{\nabla }\times \mathbf{A})\cdot (dxdy\hat{\mathbf{z}})& ={\int }_{\text{Circle of radius }R}2\pi \delta (x)\delta (y)dxdy\\ & =2\pi \end{array}$$

Which now agrees with Stokes' theorem.

(b) We start part b by writing the field in cylindrical coordinates. Let's recall the conversions:

$$\begin{array}{rl}x& =s\cos \left(\varphi \right)\\ y& =s\sin \left(\varphi \right)\end{array}$$

Further, recall

$$\begin{array}{rl}\hat{\mathbf{x}}& =\cos \varphi \hat{s}-\sin \varphi \hat{\varphi }\\ \hat{\mathbf{y}}& =\sin \varphi \hat{s}+\cos \varphi \hat{\varphi }\end{array}$$

So

$$\begin{array}{rl}\mathbf{A}& =\frac{s{\sin }^2\varphi +s{\cos }^2\varphi }{(s\cos \varphi {)}^2+(s\sin \varphi {)}^2}\hat{\varphi }\\ & =\frac{1}{s}\hat{\varphi }\end{array}$$

Which agrees with the intuition from the interactive plot above: The vector is always pointing tangentially from the circle.

The line integral will now take place using the path

$$\mathbf{r}=R\hat{s}+t\hat{\varphi }$$

With $t$ going from $0$ to $2\pi$. It is clear that the result will again be $2\pi$. For the other side of Stokes' theorem (surface integral of the curl), it is easy to see that (using the cylindrical curl) $\mathrm{\nabla }\times \mathbf{A}=\mathbf{0}$. To fix this, we again bring Dirac Deltas, now in the form:

$$\mathrm{\nabla }\times \mathbf{A}\frac{\delta (s)}{s}\hat{z}$$

And notice how this way:

$$\begin{array}{rl}{\int }_{\text{Circle of radius }R}(\mathrm{\nabla }\times \mathbf{A})\cdot d\mathbf{A}& ={\int }_{\varphi =0}^{2\pi }{\int }_0^R\frac{\delta (s)}{\cancel{s}}(\cancel{s}dsd\varphi )\\ & ={\int }_0^{2\pi }d\varphi \\ & =2\pi \end{array}$$

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