Learning Goal: How do 2 ordinary waves build up a "standing" wave? A very generic formula for a traveling wave is: y1(x,t)=Asin(
1 answer:
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Answer:
(a) 0.0178 Ω
(b) 3.4 A
(c) 6.4 x 10⁵ A/m²
(d) 9.01 x 10⁻³ V/m
Explanation:
(a)
σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹
d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m
Area of cross-section of the wire is given as
A = (0.25) π d²
A = (0.25) (3.14) (2.6 x 10⁻³)²
A = 5.3 x 10⁻⁶ m²
L = length of the wire = 6.7 m
Resistance of the wire is given as
R = 0.0178 Ω
(b)
V = potential drop across the ends of wire = 0.060 volts
i = current flowing in the wire
Using ohm's law, current flowing is given as
i = 3.4 A
(c)
Current density is given as
J = 6.4 x 10⁵ A/m²
(d)
Magnitude of electric field is given as
E = 9.01 x 10⁻³ V/m
Answer:
a) E = ρ / e0
b) E = ρ*a / (e0 * r)
c) E = 0
Explanation:
Because of the geometry, the electric field lines will all have a radial direction.
Using Gauss law
Using a Gaussian surface that is cylinder concentric to the cable, the side walls will have a flux of zero, because the electric field lines will be perpendicular. The round wall of the cylinder will have the electric field lines normal to it.
We can make this cylinder of different radii to evaluate the electric field at different points.
Then:
A = 2*π*r (area of cylinder per unit of length)
Q/e0 = 2*π*r*E
E = Q / (2*π*e0*r)
Where Q is the charge contained inside the cylinder.
Inside the cable core:
There is a uniform charge density ρ
Q(r) = ρ * 2*π*r
Then
E = ρ * 2*π*r / (2*π*e0*r)
E = ρ / e0 (electric field is constant inside the charged cylinder.
Between ther inner cilinder and the tube:
Q = ρ * 2*π*a
E = ρ * 2*π*a / (2*π*e0*r)
E = ρ*a / (e0 * r)
Outside the tube, the charges of the core cancel each other.
E=0