Physics 2220 – Module 11 Homework
01.
What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is 10 V /
m?
Use the relationship between the electric field and magnetic field for electromagnetic waves:
Emax
=c
B max
E
(10 V/m)
B max = max =
= 3.3 × 10−8 T
8
c
(3.00 × 10 m/s)
02.
A radio wave is traveling in the negative y-direction. What is the direction of E at a point where B is in
the positive x-direction?
The equation for the Poynting vector can be used to determine the direction of the electric field.
1
⃗
S = μo ( ⃗
E×⃗
B)
The Poynting vector is in the same direction as the velocity vector. Rewrite the equation to include the
specific direction of the velocity and the electric field.
1
S (− ̂j) = μ ( ⃗
E × B ̂i )
o
The unknown direction of the electric field must be able to be crossed into the i-hat vector such that it
would result in the negative j-hat direction. This can only be the negative k-hat, or negative zdirection.
1
S (− ̂j) = μ o (E (−k̂ ) × B ̂i)
03.
(a)
What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude
is 100 V / m?
E max
=c
Bmax
E
(100 V/m)
B max = max =
= 3.3 × 10−7 T
8
c
(3.00 × 10 m/s)
(b)
What is the intensity of the wave?
The intensity of the wave in terms of the energy density:
I = S av = cU av =
I=
1
2
c ϵo E max
2
1
8
−12
2
2
2
2
(3.00 × 10 m/s) (8.854 × 10 C /N m ) (100 N / C) = 13.3 W / m
2
04.
A monochromatic light source with a power output of 70.0 W radiates light of wavelength 700 nm
uniformly in all directions. Calculate the magnitude of the maximum electric and magnetic fields at a
distance of 4.00 m from the source.
Start with the relationship between intensity and the time averaged poynting vector.
S ave = I =
2
P ave
4πr
2
=
E max
2 μ0 c
Solve for the maximum electric field:
E max =
√
μ 0 cP ave
2πr
2
=
√
−7
(4 π × 10
2
8
N / A ) (3.00 × 10 m/s) (70.0 W)
= 16.2 V/m
2
2 π (4.00 m)
And now find the maximum magnetic field:
E max
=c
B max
E
(16.2 V/m)
B max = max =
= 5.4 × 10−8 T
8
c
(3.00 × 10 m/s)
05.
At what distance from a 10 W point source of electromagnetic waves is the magnetic field amplitude 1.0
μT?
Start with the equation for wave intensity:
Pav
Pav
E 2o
I=
=
=
A
4 π r 2 2c μ o
Substitute in the relationship between the electric and magnetic field strengths.
E o = cBo
I=
E2o
cB 2o
=
2c μ o 2 μ o
Solve for the distance.
2
P av
cBo
2 =
2 μo
4 πr
r=
06.
√
P av μo
2
2 π cBo
=
√
−7
2
(10 W ) (4 π × 10 N / A )
8
−6
2 = 8.2 cm
2 π (3.00 × 10 m/s) (1.0 × 10 T)
An intense light source radiates uniformly in all directions. At a distance of 6.0 m from the source, the
radiation pressure on a perfectly absorbing surface is 8.0 × 10-6 Pa. What is the total average power
output of the source?
Assuming complete absorption, the pressure is related to the average Poynting vector / Intensity by:
P=
S ave
I
=
c
c
The average power is related to the intensity by:
I=
P ave
P ave
=
A
4 π r2
Combine these two relationships and solve for the average power.
P ave
I
=
c 4 π cr 2
P ave = 4 π Pcr 2
P=
P ave = 4 π (8.0 × 10−6 Pa) (3.00 × 108 m/s) (6.0 m )2 = 1.1 × 10 6 W