Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and
2 answers:
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Answer:
1341.03 V/m
Explanation:
The power output per unit area is the intensity and also the is the magnitude of the Poynting vector.
= cε₀
⇒ = cε₀
Where;
P is the power output
A is the area of the beam
c is speed of light
ε₀ is permittivity of free space 8.85 × 10⁻¹² F/m
is the average (rms) value of electric field
Making electricfield the subject of the equation
= P / Acε₀
= √(P / Acε₀)
But area A = πr²
= √(P / πr²cε₀)
Given:
Output power, P = 15 mW = 0. 015 W
Diameter, d = 2 mm = 0.002 m
⇒ Radius,
Solving for average (rms) value of electric field;
= 1341.03 V/m
Answer:
a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W
Explanation:
This exercise is about Faraday's law
fem =
where the magnetic flux is
Ф = B x A
the bold are vectors
A = π r²
we assume that the angle between the magnetic field and the normal to the area is zero
fem = - B π 2r dr/dt = - 2π B r v
linear and angular velocity are related
v = w r
w = 2π f
v = 2π f r
we substitute
fem = - 2π B r (2π f r)
fem = -4π² B f r²
For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T
we reduce the magnitudes to the SI system
f = 2 rev / s (2π rad / 1 rev) = 4π Hz
we calculate
fem = - 4π² 428 10⁻⁶ 4π 0.10²
fem = - 16π³ 428 10⁻⁶ 0.010
fem = - 2.1514 10⁻⁴ V
for the current let's use Ohm's law
V = I R
I = V / R
I = -2.1514 10⁻⁴ / 0.00336
I = - 64.0 10⁻³ A
Electric power is
P = V I
P = 2.1514 10⁻⁴ 64.0 10⁻³
P = 1.38 10⁻⁶ W