Chapter 22, Problem 4P

To determine

The peak magnitude of electric field.

Answer to Problem 4P

Frequency is $80.0kHz$ .

The rms strength of the electric field is $2.33V/m$

The direction: horizontal north south line.

Explanation of Solution

Given info:

Rms value of Magnetic field, $B_{rms}=6.75\times {10}^{-9}T$

Factor, $c=3.00\times {10}^{8}m/s$

Frequency, $f=80.0kHz$

Formula used:

The rms strength of the electric field is,

  $E_{rms}=c{B}_{rms}$

Where,

  $B_{rms}$ is rms value of magnetic field

  $E_{rms}$ is rms value of electric field

  $c$ is a factor

Calculation:

The frequency of two fields are always be the same. So, frequency is $f=80.0kHz$ .

Now, calculate the rms strength of electric field:

Substitute the given values in formula, $E_{rms}=c{B}_{rms}$

  $E_{rms}=\left(3.00\times {10}^{8}m/s\right)\left(7.75\times {10}^{-9}T\right)=2.33V/m$

According to the question, electric field is perpendicular to both the direction of travel and also, the magnetic field.So, electric field oscillates along the horizontal north-south line.

Conclusion: The asked values are determined.