Question

The maximum magnetic field strength of an electromagnetic field is \(5.00 \times {10^{ - 6}}\;T\). Calculate the maximum electric field strength if the wave is traveling in a medium in which the speed of the wave is \(0.75{\rm{ }}c\).

Short answer

The maximum electric power available is \(1125 V/m\)

Step-by-step solution

Definition of electromagnetic wave

Examples of waves transmitted by concurrent periodic changes in the strength of the electric and magnetic fields include radio waves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

Given Information

• The magnetic field strength:\(5.00 \times {10^{ - 6}}\;T\)
• The speed of wave is: \(0.75c\)

Calculate the maximum electric field strength

The formula describing the relationship between the electric field\(\left( E \right)\), the magnetic field\(\left( B \right)\), and the speed of light\(\left( c \right)\)is:

\(E = Bc\)

Because the speed is given as\(0.75{\rm{ }}c\), we will substitute\(0.75{\rm{ }}c\)for\(c\)in the above equation. Now enter in the values of\(B\)and\(0.75{\rm{ }}c\)solve for\(E\).

\(E = Bc\)

\(\begin{aligned} &= \left( {5 \times {{10}^{ - 6}}\;T} \right) \times \left( {0.75 \times 3.00 \times {{10}^8}} \right)\left( {\frac{m}{s}} \right)\\ &= \left( {5 \times {{10}^{ - 6}}} \right) \times \left( {0.75 \times 3.00 \times {{10}^8}} \right)\left( {\frac{{V \times s}}{{{m^2}}}} \right)\left( {\frac{m}{s}} \right)\\ &= 11.25 \times {10^2}\frac{V}{m}\\ &= 1125\frac{V}{m}\end{aligned}\)

Therefore, the electric power is \(E = 1125 V/m\).