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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

Expert verified

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

Step by step solution

01

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.

02

Given Information

  • The magnetic field strength:\(5.00 \times {10^{ - 6}}\;T\)
  • The speed of wave is: \(0.75c\)
03

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 \(E = 1125 V/m\).

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