Question

What is the wavelength of the electromagnetic wave emitted by the oscillator–antenna system of the following Fig. 33-3 if$\mathit{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{253}\mathbf{}\mathit{\mu }\mathbf{\text{H}}$and$\mathit{C}\mathbf{=}\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{}\mathbf{\text{pF}}$?

Short answer

The wavelength of the electromagnetic wave emitted by the oscillator is $4.74\mathrm{m}$.

Step-by-step solution

Listing the given quantities

Inductance,$L=0.253\mu \text{H}=0.253\times {10}^{-6}\text{H}$ .

Capacitance,$C=25.0\text{pF}=25.0\times {10}^{-12}\text{F}$.

The velocity of light, $c=3\times {10}^8\text{m/s}$.

Understanding the concepts of frequency in terms of inductance and capacitance

Figure 33-3 shows an arrangement for generating a traveling electromagnetic wave in the radio region of the spectrum. It shows that an LC oscillator produces a sinusoidal current in the antenna, which generates the wave. The wavelength of electromagnetic waves emitted by an oscillator-antenna system can be calculated using the formula for frequency in terms of inductance and capacitance.

Formula:

$f=1/2\pi \sqrt{LC}$

Calculations of the wavelength of the electromagnetic wave 

To calculate, we have,

$f=1/2\pi \sqrt{LC}$ (1)

But, $f=\frac{c}{\lambda },$i.e.,

$\lambda =c/f$ (2)

From equations 1 and 2, we can write the formula for wavelength as,

$\begin{array}{rcl}\lambda & =& \frac{c}{1/2\pi \sqrt{LC}}\\ \lambda & =& 2\pi c\sqrt{LC}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{rcl}\lambda & =& 2\left(3.14\right)\left(3\times {10}^8\right)\sqrt{\left(0.253\times {10}^{-6}\right)\left(25.0\times {10}^{-12}\right)}\\ & =& 4.74\mathrm{m}\end{array}$

Thus, the wavelength of the electromagnetic wave emitted by the oscillator is $4.74\mathrm{m}$.