Question

A wave solution to Maxwell's Equations is given by E⃗ =E0xˆcos(5.0 z)cos( 1.03e11 t) where z is measured in centimeters and t in seconds.

(a) What is the wavelength (in rad/meter) of the wave?

(b) What is the index of refraction, n, of the medium?

(c) The associated magnetic field can be written as

B =B0f1(kz)f2(ωt)dˆ.

Give the expression for B0 in terms of E0,n, the speed of light in the vacuum c and natural constants.

What is the value of k in radians per meter?

What is the value of ω in radians per seconds?

Answer 1

plz help....!

Answer 2

a)0.0126
b)1.63e10
c)1.46 , (E_0*n)/c

d)... help

Answer 3

DOnny, could you give us the formula for c) first part? 1.46 did not work for me.
thanks

By the way, d) all of them 0

Answer 4

i think you meant b...

did the rest of the answers work? the formula is c/v ; c=speed of light, v= speed of what you calculated in which it would be (1.03e11)/(5*100)

Answer 5

Thanks DOnny. I mean the question for
What is the value of k in radians per meter?
5.5z
I tried your formula
(1.03e11)/(5.5*100)
=187272727
This does not work for c) first part, can you help please?

Answer 6

The rest I had already DOnny and the values are not very far apart, strangely enough, 1.46 for first part c) is not right in my case even experimenting with lower-higher values.

Thanks for your help.

Answer 7

Pls. tell K and w values and ans of Q1 all parts.

Answer 8

Have problem with part c) too, please help?

Answer 9

you're welcome. i need help with the remaining questions if you don't mind

Answer 10

oh btw k is 501.27

Answer 11

and omega is 1.03e11

Answer 12

THanks DOnny, yes me too.

Anyone for Problem 3)b)?
and Problem 1)f) B_0 for both cases.
Formula or values would helpful, thanks.

Answer 13

Thanks, now figured out DOnny!
values are not far off. Anyone for B_0 Problem 3)b?

Answer 14

flu i am stuck on the rest of them :/ you have some answers?

Answer 15

Which one DOnny?
I might able to help

Answer 16

Did anyone got B_0 correct for Problem 1) f)?

Answer 17

Please B_0!

Answer 18

For B_0 the formula seems to be :
E_0/c
However, I cannot get the result. can anybody work with this or help?

Answer 19

FLu, these are the following:

Q3 (all)
Q5 b)
Q6
Q7
Q8


i know its a lot :S

Answer 20

DOnny, values might vary you need to make some tries but they are usually around that value.

Q3:
a)0.197142857148
c)0.197142857148
b)?

Q5 b)
0
(-4*E_0)/c
E_0/c

Q 6:
a)0
0
(E_0/c)*cos(pi/1)
b)0.0269
c)((2*pi)/8)*10^-10
I think the 8*10^10 is omega value for you but in my case it was ^-10
just plug in your omega value in it should work.
Q7)a and b was in my case the exact value from the textbook, have a look with formula.
a)1.01*10^3
0.00000325
b)3.8*10^26
Q8:
Anonymous asked me that already, the values cannot be far off and in my case the n value was the answer for b) 1.5 and a) was 0.70, which was not far from alpha value 75.

Can you help me with Problem 1 f) DOnny? I still have issues with B_0.
Thanks

Answer 21

Thanks Flu!
Anybody for Question 1)f) B_0?

Answer 22

Anyone for B_0 for both cases?
Problem 3) b)?

Answer 23

my 1 f) are as follows:

-1.17e-7
4.72
-1.01e9

2.81e-7
1.567
-3.62e8

the answers you provided for Q6 a regarding wavelength are wrong in my case though. i have a value of k=1e2 and omega= 3e10

also Q3 completely wrong (formulas provided will be appreciated)

and lastly Q8 a) , with a value of n=1.55 and alpha= 84 degrees , i really dunnno where to start

Answer 24

Donny can you tell us the formula for 1)f) B_0 please?

Answer 25

How to solve B_0 please? Any formula please?

Answer 26

B_0!?

Answer 27

B_0 for Problem 3 please?

Answer 28

Pl tell formula for Q6 part b,c and Q8 n=1.45 and alpha=66

Answer 29

P,
λ = 2*pi/k
T = 2*pi/w
K and W at you are given

Answer 30

BoobleGum is this formula for Problem 3 B_0?

Answer 31

Please Problem 3 B_0?!

Answer 32

Sorry DOnny, I did not have the formula until now X10 provided below for Problem 3, maybe enoiugh time:

Part a and Part c:
d/dt (l (sin((k lambda)/4-t omega)+sin(t omega)))/k where t=2.4*10^-16; omega=3.83*10^15; k=12775810.12; lambda=4.91*10^-7;l=1

Put this value in wolframalpha and divide the result by speed of light

and for Part b: Integrate[Cos[k z - omega t], {z, 0, lambda/4}, {x, 0, l}] where t=2.2*10^-16; omega=3.83*10^15; k=12775810.12; lambda=4.91*10^-7;l=1


Do you have by any chance the formula for Problem 1 f) B_0?

Answer 33

Hurt, No, it is for Q6 part b,c.

Answer 34

sor)

Answer 35

FLU pl. tell ans for t=2.4e-16