# Light and Magnetic Field

## Homework Statement

I'm not sure if this goes in introductory physics for not, but anyways...

A light wave is traveling in glass of index 1.50. If the electric field amplitude of the wave is known to be 100 $\frac{V}{m}$, find (a) the amplitude of the magnetic field and (b) the average magnitude of the Poynting vector.

## Homework Equations

$E_{e} = \frac{power}{area}$
$E_{e} = \frac{1}{2}ε_{0}cE_{0}^{2}$
$n = \frac{c}{\upsilon}$
$E_{0} = cB_{0}$
$S = ε_{0}c^{2}E_{0}B_{0}$
$P = IV$

Where
$E_{e}$ is the irradiance
$c ≈ 2.998 X 10^{8} \frac{m}{s}$ is the speed of light
$E_{0}$ is the magnitude of the magnetic field
$ε_{0} ≈ 8.8542 X 10^{-12} \frac{(C s)^{2}}{kg m^{3}}$ is the permittivity of vacuum
n is the refractive index of a material
$\upsilon$ is the velocity of light through the material
$B_{0}$ is the magnitude of the magnetic field
S is the magnitude of the Poynting vector
P is power
V is voltage
I is current

## The Attempt at a Solution

For part (a)
I seem to be having some issues processing the given information. I know that irradiance $E_{e}$ is power $P$ divided by area$A$. I have 100 $\frac{V}{m}$, which isn't the irradiance $E_{e}$. Without being able to find the irradiance $E_{e}$ I'm not sure how to proceed. I'm unsure how to apply the knowledge of the refraction index n. I can solve for the speed of the light through the material $\upsilon$ but I'm not sure what good that really does.

$n = \frac{c}{\upsilon}$
$\upsilon = \frac{c}{n} = \frac{2.998 X 10^{8} \frac{m}{s}}{1.5} ≈ 1.999 X 10^{8} \frac{m}{s}$

Once I find the irradiance I can solve for amplitude of the electric field
$E_{e} = \frac{1}{2}ε_{0}cE_{0}^{2}$
$E_{0} = \sqrt{\frac{2E_{e}}{ε_{0}c}} = \sqrt{\frac{2E_{e}}{(2.998 X 10^{8} \frac{m}{s})(8.8542 X 10^{-12} \frac{(C s)^{2}}{kg m^{3}})}}$

Once I get this value I can solve for the amplitude of the magnetic field
$E_{0} = cB_{0}$
$B_{0} = \frac{E_{0}}{c}$

For part (b)
Once I solve part A I can solve for the average magnitude of the Poynting vector rather easily

$S = ε_{0}c^{2}E_{0}B_{0} = (8.8542 X 10^{-12} \frac{(C s)^2}{kg m^{3}})(2.998 X 10^{8} \frac{m}{s})E_{0}B_{0}$

Thanks for any help that anyone can provide me in solving this problem.

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
Simon Bridge
Science Advisor
Homework Helper
I have 100 $\frac{V}{m}$, which isn't the irradiance $E_{e}$.
You are told what it is in the problem statement -
If the electric field amplitude of the wave is known to be 100V/m
... what is the relationship between the amplitude of the electric and magnetic fields?