Question

What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is 100V/m? What is the intensity of the wave?

Answer 1

Concept and reason

The concepts required to solve the given question are intensity and the amplitude of the electromagnetic wave.

Initially, use the relation between the electric field and the magnetic field associated with an electromagnetic wave to find the magnitude of the magnetic field. Finally, calculate the intensity of the wave.

Fundamentals

The expression for the relation between the electric field and the magnetic field associated with an electromagnetic wave is as follows:

${E_0} = c{B_0}$
‎Here, ${E_0}$ is the amplitude of the electric field, c is the speed of light, and ${B_0}$ is the amplitude of the magnetic field.

The expression for the intensity of the electromagnetic wave is as follows:

$I = \frac{1}{2}c{\varepsilon _0}E_0^2$

Here, ${\varepsilon _0}$ is the permittivity of free space.

Substitute 100 V/m for ${E_0}$ and $3 \times {10^8}{\rm{ m/s}}$ for c in the equation ${E_0} = c{B_0}$.

$100{\rm{ V/m}} = \left( {3 \times {{10}^8}{\rm{ m/s}}} \right){B_0}$

Solve for ${B_0}$.

$\begin{array}{c}\\{B_0} = \frac{{100{\rm{ V/m}}}}{{3 \times {{10}^8}{\rm{ m/s}}}}\\\\ = 33.33 \times {10^{ - 8}}{\rm{ T }}\\\end{array}$

Substitute 100 V/m for ${E_0}$ and $3 \times {10^8}{\rm{ m/s}}$ for c, and $8.854 \times {10^{ - 12}}{\rm{ N}} \cdot {{\rm{m}}^{ - 1}} \cdot {{\rm{C}}^{ - 2}}$ for ${\varepsilon _0}$ in the equation $I = \frac{1}{2}c{\varepsilon _0}E_0^2$.

$\begin{array}{c}\\I = \frac{1}{2}\left( {3 \times {{10}^8}{\rm{ m/s}}} \right)\left( {8.854 \times {{10}^{ - 12}}{\rm{ N}} \cdot {{\rm{m}}^{ - 1}} \cdot {{\rm{C}}^{ - 2}}} \right){\left( {100{\rm{ V/m}}} \right)^2}\\\\ = 13.3{\rm{ W/}}{{\rm{m}}^{\rm{2}}}\\\end{array}$

Ans:

The magnitude of the amplitude of the magnetic field is $33.33 \times {10^{ - 8}}{\rm{ T}}$ and the magnitude of the intensity of the electromagnetic wave is $13.3{\rm{ W/}}{{\rm{m}}^{\rm{2}}}$.