Maximum value of the electric field..

[imatreyu]

Homework Statement

a radio transmitter broadcasts uniformly in all directions with an average power of 15 kW. Compute the maximum value of the electric field in its radio wave at the following distances from the transmitter: (a) 1 km, (b) 10 km, (c) 100 km.

Homework Equations

EmBm/2mu = Em^2/ 2muc=c/2mu Bm^2

The Attempt at a Solution

Just doing (A). . .
15000 w/ pi 1000^2 = 3e8 (Bm^2)/ 2(4pi e-7)

E/b=c
E=1.8974 N/C

Ugh I think this is totally wrong.g. . ..

 

[anathema]

Hello! Thank you for posting this question on the forum. I am happy to help you with finding the maximum value of the electric field in the radio wave emitted by the transmitter.

Firstly, let us define some variables:
- P: average power of the transmitter (15 kW)
- d: distance from the transmitter (in meters)
- E: maximum value of the electric field (in N/C)
- c: speed of light (3 x 10^8 m/s)
- μ: permeability of free space (4π x 10^-7 N/A^2)

Now, using the formula for the power of an electromagnetic wave, we can write:

P = 0.5 x ε0 x c x E^2 x A

Where ε0 is the permittivity of free space and A is the surface area of a sphere with radius d (since the transmitter broadcasts uniformly in all directions).

Solving for E, we get:

E = √(2P / ε0 x c x A)

Substituting the values, we get:

E = √(2 x 15,000 / (8.85 x 10^-12) x (3 x 10^8) x (4π x d^2))

For part (a) where d = 1 km, we get:

E = √(2 x 15,000 / (8.85 x 10^-12) x (3 x 10^8) x (4π x (1 x 10^3)^2))
= 1.92 x 10^-5 N/C

Similarly, for part (b) where d = 10 km, we get:

E = √(2 x 15,000 / (8.85 x 10^-12) x (3 x 10^8) x (4π x (10 x 10^3)^2))
= 6.08 x 10^-6 N/C

And for part (c) where d = 100 km, we get:

E = √(2 x 15,000 / (8.85 x 10^-12) x (3 x 10^8) x (4π x (100 x 10^3)^2))
= 1.92 x 10^-6 N/C

I hope this helps! Let me know if you have any further questions.