Electromagnetism, Poynting vector

[shevchenko12]

3 (a) ii) A time-dependent electric field in vacuum is given by
⃗E= E0(0, 0, sin(ky − ωt))
where E0 is a constant.
Derive an expression for the corresponding magnetic field ⃗B. [7]
Using curl E=-dB/dt
I end up with B=(E0/c)sin(ky-wt)
Show that both ⃗E and ⃗B are perpendicular to the wave vector ⃗k = (0, k, 0). [2]
Using the dot product i found that:
k.E=(0x0,0xk,0x ⃗E)=0
k.B=(0x ⃗E/c,0xk,0x0)=0
What is the Poynting vector for this wave? [4]

N=(ExB)/μ0μr

Then we get:

(1/μ0μr)((E0 ²/c)sin(ky-wt)²,0,0)
Finally giving:
(-1/μ0.μr.c)(E0²Sin(ky-wt)²)

I think the first two parts are right but have no idea if I am doing the right thing on the last part. I have used the cross product between E and B and got my final answer for part 3. Thanks for any help in advance!

 

[rude man]

3 (a) ii) A time-dependent electric field in vacuum is given by
⃗E= E0(0, 0, sin(ky − ωt))
where E0 is a constant.
Derive an expression for the corresponding magnetic field ⃗B. [7]
Using curl E=-dB/dt
I end up with B=(E0/c)sin(ky-wt)
Show that both ⃗E and ⃗B are perpendicular to the wave vector ⃗k = (0, k, 0). [2]
Using the dot product i found that:
k.E=(0x0,0xk,0x ⃗E)=0
k.B=(0x ⃗E/c,0xk,0x0)=0
What is the Poynting vector for this wave? [4]

N=(ExB)/μ0μr

Then we get:

(1/μ0μr)((E0 ²/c)sin(ky-wt)²,0,0)
Finally giving:
(-1/μ0.μr.c)(E0²Sin(ky-wt)²)

I think the first two parts are right but have no idea if I am doing the right thing on the last part. I have used the cross product between E and B and got my final answer for part 3. Thanks for any help in advance!

Basically I see nothing wrong with what you did (except for possibly a sign error.)
But I'm not sure you took the dot-products right.
The E field has a z component so the B field has an x component & obviously each of these when dotted into the y direction of propagation gives zero. Is that what you wrote?

 

[shevchenko12]

Erm, i think that's what i did. i knew that the dot products would give 0 as each one is along a different axis. Thanks for the reply btw, its good to know that I'm going along the right line.