Chapter 22, Problem 3P

To determine

The peak magnitude of electric field.

Answer to Problem 3P

The answer is $5.25N/C$

Explanation of Solution

Given info:

Magnetic field, $B=17.5nT=17.5\times {10}^{-9}T$

Factor, $c=3\times {10}^{8}m/s$

Formula used:

In anelectromagneticwave,

  $E=cB$

Where,

  $B$ is magnetic field

  $E$ magnitude of electric field

  $c$ is a factor

Calculation:

In an electromagnetic wave, magnitude of electric field is always c times that of mangnetic field.

Substitute given and known values in formula, $E=cB$

  $E=\left(3\times {10}^{8}m/s\right)\left(17.5\times {10}^{-9}T\right)=5.25N/C$

Conclusion: The peak magnitude is $5.25N/C$ .