Chapter 22, Problem 44GP

To determine

The maximum values of the electric field and the magnetic field. Also, the direction of the travelling wave.

Answer to Problem 44GP

The answers are $649\text{V/m}$ and $2.16\times {10}^{-6}\text{T}$ . The direction of the wave will be downward.

Explanation of Solution

Given data:

The rate of the energy flow, $S=560W/m^2$

Speed of the light, $c=3\times {10}^8\text{m/s}$

Permittivity of vacuum, ${\epsilon }_0=8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}$

Formula used:

It is known that: $E_0^2=\frac{S}{\left({\epsilon }_{0}c\right)}$ and ${\text{B}}_{\text{0}}=\frac{{\text{E}}_0}{c}$

Where,

  $S$ is the power flux.

  $c$ is speed of the light.

  $E_0$ is the electric field.

  ${\epsilon }_0$ is the permittivity of vacuum.

  ${\text{B}}_{\text{0}}$ is the magnetic field.

Calculation:

In order to find the direction of the wave, all we have to do is curl the finger of the right hand into the electric field (south), magnetic field (west) and our thumb show the direction of the wave that will be downward.

Substitute the given data into the formula $E_0^2=\frac{S}{\left({\epsilon }_{0}c\right)}$ to find a electric field.

  $E_0=\sqrt{\frac{S}{\left({\epsilon }_{0}c\right)}}=\sqrt{\frac{560W/m^2}{\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\times 3\times {10}^8\text{m/s}\right)}}=649\text{V/m}$

Now, calculate the magnetic field,

  ${\text{B}}_{\text{0}}=\frac{{\text{E}}_0}{c}=\frac{649\text{V/m}}{{\text{3×10}}^{\text{8}}\text{m/s}}=2.16\times {10}^{-6}\text{T}$

Conclusion: The maximum values of electric field and magnetic field are $649\text{V/m}$ and $2.16\times {10}^{-6}\text{T}$ respectively.