Question

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of fre- quency 110.0 MHz in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing-wave pattern is determined to be in its eighth harmonic, how long is the cavity?

Answer 1

The nodal planes are one-half wavelength apart.

The nodal planes of \(B\) are at \(x=\lambda / 4,3 \lambda / 4,5 \lambda / 4, \ldots\), which are \(\lambda / 2\) apart.

ExECUTE:

(a) The wavelength is \(\lambda=c / f=\left(3.000 \times 10^{8} \mathrm{~m} / \mathrm{s}\right) /\left(110.0 \times 10^{6} \mathrm{~Hz}\right)=2.727 \mathrm{~m}\). So the nodal planes are at \((2.727 \mathrm{~m}) / 2=1.364 \mathrm{~m}\) apart.

(b) For the nodal planes of \(E\), we have \(\lambda_{n}=2 L / n\), so \(L=n \lambda / 2=(8)(2.727 \mathrm{~m}) / 2=10.91 \mathrm{~m}\)