# Consider the Electric Field E(t,x,y,z) = Acos(ky-wt)k

1. Apr 16, 2013

### Smazmbazm

1. The problem statement, all variables and given/known data
Consider the Electric Field $E(t,x,y,z) = A\cos(ky-wt)\hat{k}$

a) Find the magnetic field such that $\partial_t B + \nabla \times E = 0$
b) Show that $\nabla \cdot E = 0$ and that $\nabla \cdot B = 0$.

3. The attempt at a solution

So for part a, the curl of $E = -\partial B / \partial t$

Curl of E works out to be $-Ak\sin(ky-wt)$ I think but then I'm not sure how you get the magnetic field.

For part b, div E = 0. I don't understand how to get this. I do the calculations and get $-A*t*\sin(k*y - t*w) - A*y*\sin(k*y - t*w)$? Where am I going wrong? I don't know how to do the second part at all, div B = 0.

Last edited by a moderator: Apr 16, 2013
2. Apr 16, 2013

### HallsofIvy

The curl of a vector field is a vector field. What is the direction vector?

Find B by solving the equation you are given:
$$\nabla\times \vec{E}= -\frac{\partial\vec{B}}{\partial t}$$

div fi+ gj+ hk is f_x+ g_y+ h_z. n Here f and g are 0 while h is a function only of y and t, not z.

Last edited by a moderator: Apr 16, 2013
3. Apr 16, 2013

### tiny-tim

Hi Smazmbazm!

(i assume k is the unit vector in the z direction)
in which direction?
i'm not sure how you got that

div is simply ∂/∂x of the x coordinate + … + …

try again

4. Apr 16, 2013

### Smazmbazm

Right so for the first part $\nabla \times E = [\partial / \partial y Acos(ky - wt) - 0]\hat{i} + [\partial / \partial x Acos(ky - wt) - 0]\hat{j} + [0 - 0]\hat{k}$.

$\nabla \times E = -Aksin(ky - wt)\hat{i}$, correct?

For b, right. I think I confused myself. The definition of div is $\partial f / \partial x + \partial g / \partial y + \partial h / \partial z$. Is that f, g and h referring to each direction? Like $\hat{i}, \hat{j}, \hat{k}$? If so, that makes more sense

Thanks guys

5. Apr 16, 2013

### tiny-tim

correct
yup, that's div(f,g,h)

6. Apr 16, 2013

### Smazmbazm

Sorry the previous post should be $\partial B / \partial t = -Aksin(ky - wt)\hat{i}$ rather then $\ \nabla \times E = -Aksin(ky - wt)\hat{i}$

So to get the electric field you integrate -Aksin(ky - wt) with respect to t? Not sure where to go from there..

7. Apr 16, 2013

### tiny-tim

Now I'm getting confused …

isn't it the magnetic field that you want?

Yes, integrate ∂B/∂t wrt t, and you get B.

8. Apr 16, 2013

### Smazmbazm

Yes, sorry, magnetic not electric. Ok, thanks tiny-tim