At Which the Electric Field Is Changing from Equation

Homework 9 Ch22: Q 1; P 1, 3, 19; Ch24: P 57, 59, 61

Question (Ch22):

1. The electric field in an EM wave traveling north oscillates in an east–west plane. Describe the direction of the magnetic field vector in this wave.

Solution If the direction of travel for the EM wave is north and the electric field oscillates east- west, then the magnetic field must oscillate up and down. For an EM wave, the direction of travel, the electric field, and the magnetic field must all be perpendicular to each other.

Problems (Ch22):

*1. (II) At a given instant, a 1.8-A current flows in the wires connected to a parallel- plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 1.60 cm on a side?

Solution The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from equation:

 E E E I D I D   0  I D   0 A   t t t  0 A

E 1.8A E 14  2   7.9 10 V / m  s t 8.851012 F / m1.60 102 m t

3. (I) If the magnetic field in a traveling EM wave has a peak magnitude of 17.5 nT at a given point, what is the peak magnitude of the electric field?

Solution The electric field is E cB 3.00  108 m s 17.5  10 9 T  5.25V m. *19. (II) The magnetic field in a traveling EM wave has an rms strength of 28.5 nT. How long does it take to deliver 235 J of energy to 1.00 cm 2 of a wall that it hits perpendicularly?

Solution The intensity (the average energy per unit area per unit time) is

8- 9 2 cB 2 (3.00创 10 m s)( 28.5 10 T) I =rms = = 0.194W m2 . -7 m0 (4p 10 Tg m A)

U We find the time using definition of intensity: I  At

DU (235J) 7 DU DUm Dt = = =1.21� 10 s 140 days. Dt = = 0 . AI -4 2 2 2 (1.00 10 m)( 0.194W m ) AI AcBrms

Problems (Ch24):

57. (II) At what angle should the axes of two Polaroids be placed so as to reduce the 1 1 intensity of the incident unpolarized light to (a) 3 , (b) 10 ?

Solution

If the initial intensity is I0 , through the two sheets we have

1 I1 2 I 0 , 2 I2 I 1 cos ; which means

I2 1 2  2 cos . I0

I 2 1 , 1 1 2 (a) For  3 3 2 cos  gives  35.3  . I0

I 2  1 , 1 1 2 (b) For 10 10 2 cos  gives  63.4  . I0 59. (II) Two polarizers are oriented at 38.0° to one another. Light polarized at a 19.0° angle to each polarizer passes through both. What percent reduction in intensity takes place?

Solution Through the successive sheets we have 2 I1 I 0cos 1 , 2 I2 I 1cos 2 , which gives 2 2 2 2 I2 I 0cos 1 cos  2  I 0 cos 19.0  cos 38.0   0.555 I 0 . Thus the reduction is 44.5%.

61. (II) Unpolarized light passes through five successive ideal polarizers, each of whose axis makes a 45° angle with the previous one. What fraction of the light intensity is transmitted?

Solution

1 I1  2 I 0 2 ° 1 1 I 2  I1 cos 45  2 I1  4 I 0 2 ° 1 1 I 3  I 2 cos 45  2 I 2  8 I 0 2 ° 1 1 I 4  I 3 cos 45  2 I 3  16 I 0 2 ° 1 1 I 5  I 4 cos 45  2 I 4  32 I 0