"The intensity of the field radiated at a distance r >> λ from a current element Idz is given by the magnetic filed intensity $$\hat{H} = \frac{I }{2\lambda r} sin(\theta) dz$$ Find the electric field strength in free space at a point 1km from a λ/2 dipole in a direction normal to its length when fed with a current of 0.1A (rms) at its center. Given that the radiation resistance of the λ/2 dipole is equal to 74 ohm."

I was given this question as an assignment, but I'm struggling to find an accurate solution. Could anyone outline the methods used to solve this problem?

  • \$\begingroup\$ Thanks for admitting that it is a homework assignment. Please show what attempts you have made at solving it. \$\endgroup\$
    – Aaron
    Jul 13, 2020 at 19:33
  • \$\begingroup\$ I just multiplied (H) vector by the intrinsic impedance of free space and substituted λ/2 for dz to get (E) vector. I also substituted the max. value of the current given in rms for (I) in (H) and 10^3 for (r). Then, I got an (E) expression as a function of (theta), but I'm not sure if this is correct or what I'd do with the radiation resistance value of the half-wavelength dipole. \$\endgroup\$
    – tin tan
    Jul 13, 2020 at 19:51

1 Answer 1


Some guidance: Your antenna is NOT a current element, a half-wavelength dipole must be considered to have non-uniform current distribution. The current element far-field H formula must be integrated - a highly complex task, but well presented in antenna theory books. It's complex because all current elements have different phase angles and the current distribution function isn't simple. Sinusoidal distribution is often used approximation, but it's far from accurate.

I guess you should forget current element radiation integration and use antenna handbooks to find the gain of your dipole. With radiation resistance you calculate the radiated power and find the intensity (watts/sq.meter) at the wanted distance with the gain factor. The gain becomes from the directivity, it doesn't increase energy.

From intensity you calculate E and H as rms values. Their product E*H is the radiation intensity. E and H in free air and far away from the antenna have ratio 377 Ohms. H and E are perpedicular, have the same phase angle and E is in parallel with the dipole.


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