Antennas are used for receiving and transmitting the electromagnetic signals. Their size depends upon the operating frequency / wavelength. Higher is the frequency, lower is the size of antenna. They work on Maxwell equations for field theory. They are of various types for different applications like TV transmission, AM transmission, FM transmission and satellite transmission. The waves travel in free space.
The free space wave number ‘k0’ is defined as
\(k_0=\omega_0\sqrt{\mu_0\epsilon_0}\)
Where the wave number comes from. Substituting a time-harmonic field into Maxwell's equations in a source-free lossless medium gives the wave equation
\(\nabla^{2}\mathbf{E}+\omega^{2}\mu\epsilon\,\mathbf{E}=0\)
Comparing with the standard form \(\nabla^{2}\mathbf{E}+k^{2}\mathbf{E}=0\) identifies
\(k=\omega\sqrt{\mu\epsilon}\)
and in free space, with μ0 and ε0,
\(k_0=\omega\sqrt{\mu_0\epsilon_0}\)
which is option 1.
Check it against the velocity of light. Since \(c=1/\sqrt{\mu_0\epsilon_0}\),
\(k_0=\dfrac{\omega}{c}=\dfrac{2\pi f}{c}=\dfrac{2\pi}{\lambda}\)
the familiar "phase shift per metre" form. This is the safest confirmation available: the wave number must equal 2π/λ, and only option 1 delivers that.
Check the units. k0 is a phase constant, so it must come out in radians per metre.
| Option | Expression | Units |
|---|---|---|
| 1 | \(\omega\sqrt{\mu_0\epsilon_0}=\omega/c\) | rad/m ✓ |
| 2 | \(\omega\sqrt{\mu_0/\epsilon_0}=\omega\eta_0\) | ohm-per-second — wrong |
| 3 | \(\omega/\sqrt{\mu_0\epsilon_0}=\omega c\) | m/s2 — wrong |
| 4 | \(\omega\sqrt{\mu_0/\epsilon_0}\) again | wrong |
Do not confuse k0 with η0. The two free-space constants are built from the same pair of quantities but combined differently:
\(k_0=\omega\sqrt{\mu_0\epsilon_0}=\dfrac{2\pi}{\lambda}, \qquad \eta_0=\sqrt{\dfrac{\mu_0}{\epsilon_0}}=120\pi\approx 377\ \Omega\)
The product under the root gives the wave number; the ratio gives the intrinsic impedance. Options 2 and 4 are built from the ratio and are therefore impedance-like, not phase constants.
Why the passage mentions antenna size. Because \(k_0=2\pi/\lambda\), all antenna dimensions are naturally measured in units of k0L. A half-wave dipole is the length for which \(k_0L=\pi\), and raising the frequency raises k0, so the same electrical length is achieved by a physically shorter antenna — exactly the statement made in the passage.
Hence, \(k_0=\omega_0\sqrt{\mu_0\epsilon_0}\).
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Important Questions from Electromagnetic Wave Propagation
For sky waves, following statements are given:
(A) n > 1, this shows 81 \(\rm\frac{N}{f^2}\) positive
(B) n > 1, show 81 \(\rm\frac{N}{f^2}\) Negative
(C) n < 1 shows 81 \(\rm\frac{N}{f^2}\) < 1
(D) v g x v p= c 2
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Choose the correct answer from the options given below:
If the Polarization vector is given as N and the Direction of propagation is given as K then which one of the following relations is correct?
The wave length (λ) in meters of an electromagnetic wave is related to its frequency (f) in MHz as:
Bending of light wave as it passes between material of different optical density
The wave impedance of a medium is equal to:
