SME
Electromagnetic Theory
Electromagnetic waves
Use this for “A laser beam has an intensity of 2.5 kW/m². Find the amplitudes of its electric and magnetic fields and the pressure it puts on a mirror.”
Diagram
B₀
=
E₀
c
=
1,013
V/m
÷
2.998 ×
10
8
m/s
B₀
=
3.377 ×
10
-6
T
I
=
E₀
2
2η
=
1,
013
2
2 × 376.7 Ω
=
1,361
W/
m
2
E
⟂
B
⟂
the
direction
of
travel,
in
step;
the
Poynting
vector
S
=
E
×
B
μ₀
points
along
it.
This is an example. Type your own values to start a new problem.
½cε₀
(
E₀ (V/m)
)
2
=
I (W/m²)
B₀
Magnetic field amplitude
calculated
Pₐ
Pressure on an absorber
calculated · Pa
Pᵣ
Pressure on a mirror
calculated · Pa
Formulas
I
=
1
2
×
3.00 ×
10
8
×
8.85 ×
10
-12
×
E₀
2
1,361
=
1
2
×
3.00 ×
10
8
×
8.85 ×
10
-12
×
1,
013
2
B₀
=
E₀
3.00 ×
10
8
3.375 ×
10
-6
=
1,013
3.00 ×
10
8
Pₐ
=
I
3.00 ×
10
8
4.537 ×
10
-6
=
1,361
3.00 ×
10
8
Pᵣ
=
2
×
I
3.00 ×
10
8
9.073 ×
10
-6
=
2
×
1,361
3.00 ×
10
8
Formulas are worked in T. The step-by-step shows the conversions.
Assumptions
•
A plane wave in vacuum: E ⟂ B ⟂ the direction of travel, in step, with c = 3.00 × 10⁸ m/s and ε₀ = 8.85 × 10⁻¹² F/m.
•
The intensity is the Poynting vector S = E × B ÷ μ₀ averaged over a cycle: I = ½cε₀E₀² (the average of sin² is ½).
•
The wave carries momentum I ÷ c through each square meter each second: an absorber takes it all, a mirror sends it back (twice the push).
Step-by-step
WE KNOW
I = 1,361 W/m²
FIND
electric field amplitude (E₀), magnetic field amplitude (B₀), pressure on an absorber (Pₐ), pressure on a mirror (Pᵣ)
Step 1 · Find electric field amplitude (E₀)
Use I = ½ × 3.00 × 10⁸ × 8.85 × 10⁻¹² × E₀²
Undo
I
=
1
2
cε₀
E₀
2
:
double
I
,
divide
by
cε₀,
take
the
square
root.
E₀
=
√(2
×
I
3.00 ×
10
8
× 8.85 ×
10
-12
)
E₀
=
√(2
×
1,361
3.00 ×
10
8
× 8.85 ×
10
-12
)
E₀
=
√(2
×
1,361
0.002655
)
E₀
=
√(
2,722
0.002655
)
E₀
=
√
1
,025,200
E₀
=
1,012.54
V/m
Step 2 · Find magnetic field amplitude (B₀)
Use B₀ = E₀ ÷ (3.00 × 10⁸)
Faraday’s
law
for
a
plane
wave
ties
the
two
amplitudes:
E₀
=
cB₀.
B₀
=
1,012.54
3.00 ×
10
8
B₀
=
3.3751 ×
10
-6
T
Step 3 · Find pressure on an absorber (Pₐ)
Use Pₐ = I ÷ (3.00 × 10⁸)
An
absorber
takes
all
the
momentum
that
arrives:
I
c
each
second
on
each
square
meter.
Pₐ
=
1,361
3.00 ×
10
8
Pₐ
=
4.537 ×
10
-6
Pa
Step 4 · Find pressure on a mirror (Pᵣ)
Use Pᵣ = 2 × I ÷ (3.00 × 10⁸)
A
mirror
turns
the
momentum
round,
so
the
change
is
twice
as
large:
2
I
c
.
Pᵣ
=
2
×
1,361
3.00 ×
10
8
Pᵣ
=
2,722
3.00 ×
10
8
Pᵣ
=
9.073 ×
10
-6
Pa
Step 5 · Convert the answers
Change each answer back to the unit you chose.
B₀ = 3.375 × 10⁻⁶ T = 3.375 μT (1 T = 10⁶ μT)
Check
Put the numbers back in. Both sides should match.
1,361
=
1
2
×
3.00 ×
10
8
×
8.85 ×
10
-12
×
1,
012.54
2
✓
3.375 ×
10
-6
=
1,012.54
3.00 ×
10
8
✓
4.537 ×
10
-6
=
1,361
3.00 ×
10
8
✓
9.073 ×
10
-6
=
2
×
1,361
3.00 ×
10
8
✓
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A point source: intensity and field at a distance
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