Question

In Figure 23.116 electromagnetic radiation is moving to the right, and at this time and place the electric field is horizontal and points out of the page. The magnitude of the electric field is $\mathit{E}\mathbf{=}\mathbf{3000}\mathbf{ }\mathbf{\text{N/C}}$. What are the magnitude and direction of the associated magnetic field at this time and place?

Short answer

The magnitude and direction of the associated magnetic field at this time and place are$1\times {10}^{-5} \text{T}$ and negative y direction

Step-by-step solution

Identification of given data

Magnitude of electric field$E=3000 \text{N/C}$

Concept of Ampere’s law

It states that the pulse of the electric field and magnetic field is travelling through space with speed of light. It is given as

$E=cB$ …(i)

Where,$E$ is the electric field,$B$ is the magnetic field and$c$ is the speed of light

Determining the magnitude and direction of the associated magnetic field at this time and place

By rearranging equation (i)

$B=\frac{E}{c}$

Substitute all the values in above equation

$$\begin{gathered} B=\frac{3000 \text{N/C}}{3\times {10}^8\text{m/s}} \\ =1\times {10}^{-5} \text{T} \end{gathered}$$

Direction of propagation of electromagnetic wave is given by,

$\overrightarrow{E}\times \overrightarrow{B}$

By using the right hand rule direction or magnetic field is in the negative y direction

Hence the magnitude and direction of the associated magnetic field at this time and place are$1\times {10}^{-5} \text{T}$ and negative y direction