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

[Smazmbazm]

Homework Statement

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 $\nabla \cdot B = 0$.

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.

Thanks in advanced

 

[HallsofIvy]

Homework Statement

Consider the Electric Field E(t,x,y,z) = Acos(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.

The Attempt at a Solution

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

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

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}

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.

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.

Thanks in advanced

 

[tiny-tim]

Hi Smazmbazm!

(i assume k is the unit vector in the z direction)

Curl of E works out to be -Aksin(ky-wt)

in which direction?

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)?

i'm not sure how you got that

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

try again ​

 

[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

 

[tiny-tim]

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

correct

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}?

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

 

[Smazmbazm]

Sorry the previous post should be \partial B / \partial t = -Aksin(ky - wt)\hat{i} rather then \<br /> \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..

 

[tiny-tim]

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

Now I'm getting confused …

isn't it the magnetic field that you want?

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

 

[Smazmbazm]

isn't it the magnetic field that you want?

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