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Appreciate any help you can give.

  • Thread starter madd_bm
  • Start date
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1. Homework Statement
The average intensity of sunlight reaching the earth is 1390 W / m2. A charge of 1.60 10-8 C is placed in the path of this electromagnetic wave.
(a) What is the magnitude of the maximum electric force that the charge experiences?
1.637 N


(b) If the charge is moving at a speed of 6.00 104 m/s, what is the magnitude of the maximum magnetic force that the charge could experience?



I got part A, but I cannot figure out what to do next:

Savg - inensity
c - speed of light 3e8
ε0 - permittivity of free space 8.854e-12
Erms

formula: Erms = √(Savg / c * ε0) = 723.3975, Erms = 1/√2 * E0
Erms / .707 = 1023.038
E0 = 1023.038
E0 * q = F
F = 1.637e-5

am I suppossed to muliply by the speed? Cause it shows a wrong answer when i do that. I cannot seem to find the right formula. I am just not sure....?

Thanks.
 

alphysicist

Homework Helper
2,238
1
Hi madd_bm,

1. Homework Statement
The average intensity of sunlight reaching the earth is 1390 W / m2. A charge of 1.60 10-8 C is placed in the path of this electromagnetic wave.
(a) What is the magnitude of the maximum electric force that the charge experiences?
1.637 N


(b) If the charge is moving at a speed of 6.00 104 m/s, what is the magnitude of the maximum magnetic force that the charge could experience?



I got part A, but I cannot figure out what to do next:

Savg - inensity
c - speed of light 3e8
ε0 - permittivity of free space 8.854e-12
Erms

formula: Erms = √(Savg / c * ε0) = 723.3975, Erms = 1/√2 * E0
Erms / .707 = 1023.038
E0 = 1023.038
E0 * q = F
F = 1.637e-5
For part b they want to find the magnetic force, not the electric force. What is the formula for the magnetic force on a particle? (It will have the velocity in the formula.)
 
6
0
o.k. this is what I came up with.

F = qvB

u = 1/u0 x B^2 (u0 being the permittivity of free space 4pie-7)

since u is the same as the "u" in the first equation, I used 1390 = 1/u0 x B^2 to calculate B to be 4.179e-2

then multiplied B by 1.6e-8 x 6e4 (velocity x charge) to get FB = 4.012e-5

...but this is still wrong. ?????
what am I missing?

thanks again. Whatever it comes out to be I won't know if it is right until after tomorrow. I have exceeded my maximum number of tries on the question

thanks.


edit: I just thought of this and came up with a different answer:

if E = cB, then B = 3.41e-6 (c being the speed of light 3e8, E being the Electric field from part A, 1023.038)
F = qvb, then F = 3.274e-9

this might also be correct, but like I said, I will not know until tomorrow. thanks again for your help.

another edit: wait. I think I know what I did wrong. above, I used 1390 as u. I just realized this is the average intensity, not the total.
I redid the first equation (u = 1/u0 x B^2) above using the total (9.267e-6), I then followed the same steps u = 1/u0 x B^2 to calculate B using the corrected u and then set that to F= qvB and got the same answer as in the second equation: 3.27e-9
so this is likely the correct answer. I just have to wait and see....
 
Last edited:
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Marked! I will come back to check this soon!thanks a lot.:-)
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