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Chapter 31: Q. 43 (page 903)

a. Show that $u_{E}$ and $u_{B}$ , the energy densities of the electric and magnetic fields, are equal to each other in an electromagnetic wave. In other words, show that the wave鈥檚 energy is divided equally between the electric field and the magnetic field.
b. What is the total energy density in an electromagnetic wave of intensity $1000 W / m^{2}$ ?

Short Answer

Expert verified

(a) The total energy $u$ of $E M$ wave is $u = 蔚_{o} E^{2}$

(b) The total energy density in an electromagnetic wave of intensity is $u = 3.33 脳 10^{- 6}$

Step by step solution

01

Final the energy on EM waves (part a)

Given:

Magnetic energy density $u_{B}$ is,

$u_{B} = \frac{B^{2}}{2 渭_{o}}$

Magnetic energy density $u_{E}$ is,

$u_{E} = \frac{1}{2} 蔚_{o} E^{2}$

02

Find the energy on EM waves (part a)

$E$ and $B$ are connected in $E M$ waves as $E = c B$ , where is $c = \frac{1}{\sqrt{渭_{o} 蔚_{o}}}$

We get,

$u_{E} = \frac{1}{2} 蔚_{o} E^{2} = \frac{1}{2} 蔚_{o} B^{2} c^{2} = \frac{1}{2} 蔚_{o} B^{2} \frac{1}{渭_{o} 蔚_{o}} = u_{B}$

Because the energy densities of the two waves are equal, the energy density $u$ of the $E M$ waves equals

$u = u_{E} + u_{B}$

03

The electrical density on electromagnetic wave of intensity (part b)

Given:

Intensity, $I = 1000 W / m^{2}$ and then,

$I = c 蔚_{o} E^{2}$

$E^{2} = \frac{I}{c 蔚_{o}}$

$E = \sqrt{\frac{I}{c 蔚_{o}}}$

$E = \sqrt{\frac{1000}{(3 脳 10^{8}) (8.85 脳 10^{- 12})}}$

$E = 613.7$

Then we find total energy density:

$u = 蔚_{o} E^{2}$

Substitute the values,

$u = (8.85 脳 10^{- 12}) (613.7)^{2}$

$u = 3.33 脳 10^{- 6}$

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