Compute Vector Potentials for Two-Dimensional Fields | Homework Question

[EngageEngage]

Homework Statement

Compute the vector potentials for the following two-dimensional fields:

$$\vec{F}$$ = $$\frac{xy\vec{i}}{1}$$ - $$\frac{y^{2}\vec{j}}{2}$$

Homework Equations

(1)$$\int^{1}_{0} t \vec{F}\times \frac{d\vec{r}}{dt} dt$$

The Attempt at a Solution

I solved this problem a different way, but my problem is using the above formula. The way that i did it was as follows:

$$\vec{k} \times$$$$\vec{F}$$ = $$\nabla\chi$$ = $$\left\langle\frac{y^{2}}{2},xy,0\right\rangle$$

so $$\chi$$ = $$\frac{xy^{2}}{2}$$

but, $$\vec{G} = \vec{k}\chi=\frac{xy^{2}}{2}\vec{k}$$
where $$\vec{F} = \nabla \times \vec{G}$$

This is the correct answer given by the book, with G being the vector potential. However, I cannot get to this answer using the formula (1) that I gave above. I did the following:

$$\vec{r}=t\left\langle x,y,z\right\rangle$$
$$\frac{d\vec{r}}{dt}=\left\langle x,y,z\right\rangle$$
$$\vec{F}(\vec{r}(t)) = \left\langle xyt^{2},\frac{t^{2}y^{2}}{2},0\right\rangle$$

Then, following through with the computations I get an incorrect answer. Did i set this up incorrectly? Any help at all would be greatly appreciated

 

[mighty2000]

.

Thank you for your post. I have reviewed your attempt at solving the problem and I believe you have set up the formula (1) incorrectly. The formula is used to find the vector potential for a two-dimensional field, which means that the z-component should not be included. Additionally, the vector potential is defined as a function of x and y, not t. Therefore, the correct setup for formula (1) should be:

\int^{1}_{0} t \vec{F}\times \frac{d\vec{r}}{dt} dt = \int^{1}_{0} t\left\langle xyt^{2},\frac{t^{2}y^{2}}{2}\right\rangle \times \left\langle x,y\right\rangle dt = \int^{1}_{0} \left\langle xy^{2}t^{3},\frac{xy^{2}t^{2}}{2}\right\rangle dt

Solving this integral will give you the correct vector potential, which is \vec{G} = \frac{xy^{2}}{2}\vec{k}.

I hope this helps. If you have any further questions, please do not hesitate to ask.