# Video: Electric Field and Magnetic Field Strength Relationship for Electromagnetic Waves

A certain 60.0-Hz ac powerline radiates an electromagnetic wave having a maximum electric field strength of 13.0 kV/m. What is the wavelength of this very-low-frequency electromagnetic wave? What is the magnitude of the strongest magnetic field present in this wave?

02:50

### Video Transcript

A certain 60.0-hertz ac powerline radiates an electromagnetic wave having a maximum electric field strength of 13.0 kilovolts per meter. What is the wavelength of this very-low-frequency electromagnetic wave? What is the magnitude of the strongest magnetic field present in this wave?

Weโre told in this statement that the powerline has a frequency of 60.0 hertz; weโll call this value ๐. Weโre also told that the wave it radiates has a maximum electric field strength of 13.0 kilovolts per meter; weโll call this ๐ธ sub max. We want to know the wavelength of this wave, which weโll call ๐, and we want to know the magnitude of the strongest magnetic field present in the wave, which weโll call ๐ต sub max.

To begin our solution, we can recall that, in general, the speed of a wave ๐ฃ is equal to the frequency of that wave times its wavelength. When we apply this relationship to our scenario, since we have an electromagnetic wave, our wave speed is ๐, the speed of light, which is equal to ๐ times ๐. When we rearrange to solve for ๐, the wavelength, we see that itโs equal to ๐ over ๐.

Weโve been told ๐ in the problem statement, and weโll assume that ๐ is exactly 3.00 times 10 to the eighth meters per second. We can then plug in for ๐ and ๐. And when we calculate this fraction, we find that ๐ is equal to 5.00 times 10 to the sixth meters. Thatโs the wavelength of this electromagnetic wave.

Next, we move on to solving for the maximum magnetic field strength of this wave. And to do that, we can recall a relationship between electric and magnetic fields. Given an electric field ๐ธ, the magnitude of the corresponding magnetic field ๐ต is equal to ๐ธ divided by the speed of light ๐. So in our case, ๐ต sub max equals ๐ธ sub max divided by ๐.

Plugging in for these values and making sure to write ๐ธ sub max in units of volts per meter, when we calculate this fraction, we find that ๐ต sub max is equal to 4.33 times 10 to the negative fifth tesla. Thatโs the maximum magnetic field magnitude produced by this electromagnetic wave.