- #1
Pepsi24chevy
- 65
- 0
Sry, i think i put this in the wrong forum last time.
Here is the problem i have been working on.
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.20 kW. A dipole receiving antenna 70.0 cm long is at a location 4.00 miles from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.
I got P= 3200 W
L antenna = .40m
I did A= 4pir^2 in which i get 6437.376*4pi =520748007.7
Next i know that S= energy/(area*time) but power is equal to energy/time, so I did 3200w/520748007.7 and i get 6.145006706e-6
From here i Know s= Emax/2mu_oC so i did Emax = S* 2mu_oC in which i get 6.145006706e-68 *(3*10^8)(4pi*10^-7)*2 and i get .0046332259. Now i know that Vmax= Emax(L) so i multiply that answer by .70 and i get . .00324 v or 3.24mv, but appearantly this is wrong... Where did i mess up?
Here is the problem i have been working on.
An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.20 kW. A dipole receiving antenna 70.0 cm long is at a location 4.00 miles from the transmitter. Compute the amplitude of the emf that is induced by this signal between the ends of the receiving antenna.
I got P= 3200 W
L antenna = .40m
I did A= 4pir^2 in which i get 6437.376*4pi =520748007.7
Next i know that S= energy/(area*time) but power is equal to energy/time, so I did 3200w/520748007.7 and i get 6.145006706e-6
From here i Know s= Emax/2mu_oC so i did Emax = S* 2mu_oC in which i get 6.145006706e-68 *(3*10^8)(4pi*10^-7)*2 and i get .0046332259. Now i know that Vmax= Emax(L) so i multiply that answer by .70 and i get . .00324 v or 3.24mv, but appearantly this is wrong... Where did i mess up?