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Chapter 33: Q103P (page 1009)

The rms value of the electric field in a certain light waveis $0.200 鈥� V / m$ .What is the amplitude of the associated magnetic field?

Short Answer

Expert verified

The amplitude of the associated magnetic field is $B_{a m p} = 9.43 脳 10^{- 10} T$

Step by step solution

01

Listing the given quantities

$E_{r m s} = 0.200 V / m$

02

Understanding the concepts of magnetic field

By using the relation between the magnitude of the electric field andthe magnitude of the magnetic field, given by equation $33 - 5$ ,we can find the magnetic field.

Formula:

$\frac{E}{B} = C$

$B_{r m s} = \frac{B_{a m p}}{\sqrt{2}}$

03

Calculations of theamplitude of the associated magnetic field

By using the equation, $33 鈭� 5$ we have,

$\frac{E}{B} = C$

where C is the velocity of the electromagnetic wave, which is $3 脳 10^{8} 鈥� m / s$ .

Rearranging the equation, we can find the rms magnetic field,

$\begin{matrix} B_{r m s} = \frac{E_{r m s}}{C} \\ = \frac{0.200}{3 脳 10^{8}} \\ = 6.67 脳 10^{- 10} T \end{matrix}$

This is the value of the magnetic field. To find the amplitude of the magnetic field, we can multiply this value by $\sqrt{2}$ .

$\begin{matrix} B_{r m s} = \frac{B_{a m p}}{\sqrt{2}} \\ B_{a m p} = B_{r m s} \sqrt{2} \\ = 6.67 脳 10^{鈭� 10} 脳 \sqrt{2} \\ = 9.43 脳 10^{- 10} T \end{matrix}$

Hence, the amplitude of the associated magnetic field is $B_{a m p} = 9.43 脳 10^{- 10} T$

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Most popular questions from this chapter

In Fig. 33-73, a long, straight copper wire (diameter $2 .50 鈥尘尘$ and resistance $1 .00 鈥坝﹑别谤 300 鈥尘$ ) carries a uniform current of $25 鈥碱$ in the positive x direction. For point P on the wire鈥檚 surface, calculate the magnitudes of (a) the electricfield, $\overset{鈫�}{E}$ (b) the magnetic field , $\overset{鈫�}{B}$ and (c) the Poynting vector $\overset{鈫�}{S}$ , and (d) determine the direction of $\overset{鈫�}{S}$ .

Figure 33-32 shows four long horizontal layers 础鈥揇 of different materials, with air above and below them. The index of refraction of each material is given. Rays of light are sent into the left end of each layer as shown. In which layer is there the possibility of totally trapping the light in that layer so that, after many reflections, all the light reaches the right end of the layer?

In Fig. 33-41, a beam of light, with intensity ${43鈥塛/尘}^{2}$ and polarization parallel to a y-axis, is sent into a system of two polarizing sheets with polarizing directions at angles of $胃_{1} =70掳$ and $胃_{2} =90掳$ to the y axis. What is the intensity of the light transmitted by the two-sheet system?

Unpolarized light of intensity ${10鈥尘W/m}^{2}$ is sent into a polarizing sheet as in Fig.33-11. What are

(a) the amplitude of the electric field component of the transmitted light and

(b) the radiation pressure on the sheet due to its absorbing some of the light?

The magnetic component of an electromagnetic wave in vacuum has an amplitude of $85 .8 鈥塶罢$ and an angular wave number of $4.00 m^{- 1} .$ What are (a) the frequency of the wave, (b) the rms value of the electric component, and(c) the intensity of the light?

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