Answer:
hm= 18.2 m
Explanation:
Solution is attached
Hi, please just say the Letter (A,B,C,D,E)
1 answer:
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Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
= - L dI / dt
Let's use Faraday's law
E = - d Ф_B / dt
in the case of the coil this voltage is the same, so we can equal the two relationships
- d Ф_B / dt = - L dI / dt
The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil
n d Ф_B = L dI
we can remove the differentials
n Ф_B = L I
magnetic flux is defined by
Ф_B = B . A
in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product
n B A = L I
the loop area is
A = π R²
we substitute
n B π R² = L I (1)
To find the magnetic field in the coil let's use Ampere's law
∫ B. ds = μ₀ I
where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil
s = 2π R
we solve
B 2ππ R = μ₀ I
B = μ₀ I / 2πR
we substitute in
n ( μ₀ I / 2πR) π R² = L I
n μ₀ R / 2 = L I
L = μ₀ n r / 2I
Answer:
Explanation:
The capacitance of a parallel plate capacitor is given by:
(1)
where
is the vacuum permittivity
A is the area of the plates
d is the separation between the plates
The charge stored on the capacitor is given by
(2)
where C is the capacitance and V is the voltage across the capacitor.
The displacement current in the capacitor is given by
(3)
where t is the time elapsed
Substituting (1) and (2) into (3), we find an expression for the displacement current:
where we have
Substituting into the equation, we find