Respuesta :
Answer:
L = μ₀ n r / 2I
Explanation:
This exercise we must relate several equations, let's start writing the voltage in a coil
[tex]E_{L}[/tex] = - 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
The self-inductance of the solenoid is determined as [tex]\L = \frac{\mu_o N^2 \pi R^2}{l}[/tex].
The emf induced in a solenoid is given by Faradays law;
[tex]emf = \frac{d\phi}{dt} \\\\emf = \frac{Ldi}{dt} \\\\d\phi = Ldi\\\\\phi = BA\\\\[/tex]
For "N" number of turns in the solenoid, we will have total flux;
[tex]NBA = LI\\\\\phi = LI[/tex]
Magnetic field at any point on a long solenoid is given as;
[tex]B = \frac{\mu_0 NI}{l}[/tex]
The total magnetic flux is given as;
[tex]\phi = NBA\\\\\phi = N \times \frac{\mu_0 NI}{l} \times A\\\\\phi = \frac{\mu_o N^2 IA}{l}[/tex]
[tex]Recall, LI = \phi\\\\LI = \frac{\mu_o N^2 I A}{l} \\\\L = \frac{\mu_o N^2 A}{l}\\\\L = \frac{\mu_o N^2 \pi R^2}{l}[/tex]
Thus, the self-inductance of the solenoid is determined as [tex]\L = \frac{\mu_o N^2 \pi R^2}{l}[/tex]
Learn more here:https://brainly.com/question/15612168
