Chapter 22, Problem 42GP

To determine

The values of ${\text{E}}_0$ and ${\text{B}}_{\text{0}}$ .

Answer to Problem 42GP

The answers are $\text{37}\text{.7}\text{V/m}$ and $1.26\times {10}^{-7}\text{T}$ .

Explanation of Solution

Given data:

Power, $P_0=95W$

Radius, $\text{r}=\text{2}\text{m}$

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: $E_0^2=\frac{2P_0}{\left(4\pi r^2\right)\left({\epsilon }_{0}c\right)}$ and ${\text{B}}_{\text{0}}=\frac{{\text{E}}_0}{c}$

Where,

  $P$ is the power.

  $c$ is speed of the light.

  $E$ is the electric field.

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

  $\text{r}$ is the radius.

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

Calculation:

Substitute the given data into the formula $E_0^2=\frac{2P_0}{\left(4\pi r^2\right)\left({\epsilon }_{0}c\right)}$ to find a electric field.

  $E_0=\sqrt{\frac{2P_0}{\left(4\pi r^2\right)\left({\epsilon }_{0}c\right)}}=\sqrt{\frac{2\times 95W}{4\pi {\left(2.00m\right)}^2\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}\text{/N}{\text{.m}}^{\text{2}}\times 3\times {10}^8\text{m/s}\right)}}=\text{37}\text{.7}\text{V/m}$