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Physics 112
Homework7 (Ch22 & 24)
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.
2. 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:
ID  0
 E

t
ID  0 A
E
1.8 A

t
8.85  10 12 F / m 1.60  10 2 m


I
E
 D
t  0 A
E

t

2
E
 7.9  1014 V / m  s
t

3. 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.
4. 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 cm2 of a wall that it hits perpendicularly?
Solution
The intensity (the average energy per unit area per unit time) is
I 
cBrms 2
0
3.00  10 m s  28.5  10 T 

 4  10 T m A 
9
8
7
2
 0.194 W m2 .
We find the time using definition of intensity:
U
 235J 
U
I
 t 

 1.21  107 s  140 days.
4
AI 1.00  10 m 2  0.194 W m 2 
At
t 
U 0
U

AI
AcBrms 2
Physics 112
Homework7 (Ch22 & 24)
5. At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the
incident unpolarized light to (a) 13 , (b) 101 ?
Solution
If the initial intensity is I 0 , through the two sheets we have
I1  12 I 0 ,
I 2  I1 cos2  ; which means
I2 1
 cos2  .
I0 2
(a) For
I2 1
 ,
I0 3
1
3
(b) For
I2 1
 ,
I 0 10
1
10
 12 cos 2  gives   35.3.
 12 cos2  gives   63.4.
6. 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
I1  I 0 cos2 1 ,
I 2  I1 cos2  2 , which gives
I 2  I 0 cos2 1 cos2  2  I 0  cos2 19.0  cos 2 38.0   0.555I 0 .
Thus the reduction is 44.5%.
7. 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
I 1  12 I 0
I 2  I 1 cos 2 45   12 I 1  14 I 0
I 3  I 2 cos 2 45   12 I 2  18 I 0
I 4  I 3 cos 2 45   12 I 3  161 I 0
I 5  I 4 cos 2 45   12 I 4 
1
32
I0