Question

A certain 50.0-Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 13.0 kV/m. (a) What is the wavelength of this very low frequency electromagnetic wave? (b) What is its maximum magnetic field strength?

Short answer

(a) The wavelength of the given low frequency electromagnetic wave.is$6\times {10}^{6}m$

(b) The maximum magnetic field strength of the electromagnetic wave is$4.33\times {10}^{-5}T$

Step-by-step solution

Definition of electromagnetic wave

Electromagnetic waves are those in which the electric and magnetic field vectors change sinusoidally and are perpendicular to each other as well as perpendicular to the wave's propagation direction.

Given information and Formula to be used

Frequency of the electromagnetic wave ia$f=50.0Hz$

Strength of the electric field,$E=13.0kV/m\left(\frac{1000V/m}{1kV/M}\right)=1.30\times {10}^{4}V/m$

The relation between the wavelength and frequency can be expressed as,

$\lambda =\frac{c}{f}$……………(1)

Here is the wavelength, c is the speed of light, and f is the frequency.

The relation between the magnetic field and electric field can be expressed as,

$B_0=\frac{E_0}{c}$………………(2)

Here $B_0$is the strength of magnetic field, $E_0$is the strength of electric field, and c is the speed of light.

Calculate the wavelength of the given low frequency electromagnetic wave (a)

Substitute the given data in above expression (1), we get,

$$\begin{gathered} \lambda =\frac{3\times {10}^{8}m/s}{50.0Hz} \\ =6\times {10}^{6}m \end{gathered}$$

Therefore, the wavelength of the given low frequency electromagnetic wave.is$6\times {10}^{6}m$

Calculate the maximum magnetic field strength of the electromagnetic wave (b)

Substitute the given data in above expression (2), we get,

$$\begin{gathered} B_0=\frac{13.0\times {10}^{}V/m}{3\times {10}^{8}m/s} \\ =4.33\times {10}^{-5}T \end{gathered}$$

Therefore, the maximum magnetic field strength of the electromagnetic wave is$4.33\times {10}^{-5}T$