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Chapter 33: Q10P (page 1001)

A plane electromagnetic wave has a maximum electric field magnitude of3.20×10-4V/m. Find the magnetic field amplitude.

Short Answer

Expert verified

The magnetic field amplitude is 1.067×10-12T.

Step by step solution

01

Listing the given quantities

Maximum electric field magnitude Em=3.20×10-4V/m.

02

Understanding the concepts of magnetic field

The electric Emand magnetic fields Bmare related to the wave speed c as,

c=EmBm

03

Calculations of the magnetic field amplitude

The electric and the magnetic field are related as,

c=EmBm

Where c is the wave speed c=3.0×108m/s.

Rearranging the above expression as,

Bm=Emc

Substitute the values in the above expression, and we get,

Bm=3.20×10-4V/m3.0×108m/s=1.067×10-12T

Thus, the magnetic field amplitude is 1.067×10-12T.

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Most popular questions from this chapter

In Fig. 33-70, unpolarized light is sent into the system of three polarizing sheets, where the polarizing directions of the first and third sheets are at angles θ1=30°(counter-clockwise) andθ3=30°(clockwise). What fraction of the initial light intensity emerges from the system?

A black, totally absorbing piece of cardboard of area A = 2.0cm2intercepts light with an intensity of110W/m2 from a camera strobe light. What radiation pressure is produced on the cardboard by the light?

In Fig. 33-29, unpolarized light is sent into a system of five polarizing sheets. Their polarizing directions, measured counterclockwise from the positive direction of the y axis, are the following: sheet 1,35°; sheet 2, 0°; sheet 3,0°; sheet 4,110°; sheet 5,45°. Sheet 3 is then rotated180°,counterclockwise about the light ray. During that rotation, at what angles (measured counterclockwise from the y axis) is the transmission of light through the system eliminated?

An isotropic point source emits light at wavelength500nm, at the rate of200W. A light detector is positioned400mfrom the source. What is the maximum rate∂B/∂t at which the magnetic component of the light changes with time at the detector’s location?

In Figure

(a), a beam of light in a material1is incident on a boundary at an angle θ1=40°. Some of the light travels through the material 2, and then some of it emerges into the material 3. The two boundaries between the three materials are parallel. The final direction of the beam depends, in part, on the index of refraction n3of the third material. Figure (b) gives the angle of refraction θ3in that material versus n3a range of possiblen3values. The vertical axis scale is set byθ3a=30.0° and θ3b=50.0°.(a) What is the indexof refraction of material , or is the index impossible to calculate without more information?

(b) What is the index of refraction of material 2, or is the index impossible to calculate without more information?

(c) It θ1is changed to 70°and the index of refraction of a material 3 is2.4 , what is θ3?

Figure:
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