The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoid? (b) What is the induced emf in the solenoid if its current drops from 3.2 A to 0 in 55 ms?

Respuesta :

Answer: 0.43 V

Explanation:

L = [μ(0) * N² * A] / l

Where

L = Inductance of the solenoid

N = the number of turns in the solenoid

A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

1.752*10^-3 = 0.255 * A

A = 1.752*10^-3 / 0.255

A = 0.00687 m²

A = 6.87*10^-3 m²

emf = -N(ΔΦ/Δt).........1

L = N(ΔΦ/ΔI) so that,

N*ΔΦ = ΔI*L

Substituting this in eqn 1, we have

emf = - ΔI*L / Δt

emf = - [(0 - 3.2) * 7.3*10^-3] / 55*10^-3

emf = 0.0234 / 0.055

emf = 0.43 V

The cross-sectional area of the solenoid is 6.88 x 10⁻³ m².

The induced emf in the solenoid when the current drops is 0.425 V.

The given parameters;

  • inductance of the solenoid, L = 7.3 mH
  • number of turns of the solenoid, N = 450
  • length of the solenoid, L = 24 cm = 0.24 m

The cross-sectional area of the solenoid is calculated as follows;

[tex]L = \frac{\mu_o N^2 A}{l} \\\\A = \frac{Ll}{\mu_0 N^2} \\\\A = \frac{(7.3\times 10^{-3}) \times (0.24)}{4\pi \times 10^{-7} \times 450^2} \\\\A = 6.88\times 10^{-3} \ m^2[/tex]

The induced emf in the solenoid is calculated as follows;

[tex]emf = L \frac{dI}{dt} \\\\emf = (7.3\times 10^{-3})\times \frac{(3.2 -0)}{55\times 10^{-3}} \\\\emf = 0.425 \ V[/tex]

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