. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt
1 answer:
Answer:
a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J
Explanation:
a) The inductance is a solenoid this given carrier
L =
The magnetic field inside the solenoid is
B = μ₀
hence the magnetic flux
Ф_B = B. A = μ₀
we substitute in the expression of inductance
L = N² μ₀ A /l
let's find the area of each turn
A = π r²
A = π 0.02²
A = 1.2566 10⁻³ m²
let's calculate
L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3
L = 3.29 10⁻⁴ H
b) The stored energy is
U = ½ L i²
let's calculate
U = ½ 3.29 10⁻⁴ 18²
U = 5.33 10⁻² J
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a) E = 0
b)
Explanation:
a)
We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:
where
E is the electric field
q is the charge contained by the Gaussian surface
is the vacuum permittivity
Here we want to find the electric field at a distance of
r = 12 cm = 0.12 m
Here we are between the inner radius and the outer radius of the shell:
However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.
This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:
q = 0
Therefore, the magnitude of the electric field is also zero:
E = 0
b)
Here we want to find the magnitude of the electric field at a distance of
r = 20 cm = 0.20 m
from the centre of the shell.
Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.
Therefore, it is given by:
where in this problem:
is the charge on the shell
is the distance from the centre of the shell
Substituting, we find:
Answer:
Explanation:
Let the angle between the first polariser and the second polariser axis is θ.
By using of law of Malus
(a)
Let the intensity of light coming out from the first polariser is I'
.... (1)
Now the angle between the transmission axis of the second and the third polariser is 90 - θ. Let the intensity of light coming out from the third polariser is I''.
By the law of Malus
So,
(b)
Now differentiate with respect to θ.