stephaniechavez313
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A solenoid with 465 turns has a length of 6.50 cm and a cross-sectional area of 2.90 ✕ 10⁻⁹m2. Find the solenoid's inductance and the average emf around the solenoid if the current changes from +3.50 A to −3.50 A in 5.33 ✕ 10⁻³ s. Apply the expression for the self-inductance of a solenoid and the self-induced emf of an inductor.

(a) The solenoid's inductance (in H):
_________H

(b) the average emf around the solenoid (in V):
____________V

Respuesta :

onyebuchinnaji

(a) The solenoid's inductance  is  1.212 x 10⁻⁸ H.

(b) The average emf around the solenoid is  1.59 x 10⁻⁵ V.

Inductance of the solenoid

The solenoid's inductance is calculated as follows;

L = N²μA/I

where;

  • N is number of turns of the solenoid
  • μ is permeability of free space
  • A is area of the solenoid
  • I is length of the solenoid

Substitute the given parameters and solve for Inductance of the solenoid.

L = (465² x 4π x 10⁻⁷ x 2.9 x 10⁻⁹) / (0.065)

L = 1.212 x 10⁻⁸ H

Average emf of the solenoid

emf = LdI/dt

emf = L(I₁ - I₂)/t

emf =  1.212 x 10⁻⁸(3.5 - - 3.5)/5.33 x 10⁻³

emf =  1.212 x 10⁻⁸(7 / 5.33 x 10⁻³)

emf = 1.59 x 10⁻⁵ V

Thus, the solenoid's inductance  is  1.212 x 10⁻⁸ H and the average emf around the solenoid is  1.59 x 10⁻⁵ V.

Learn more about inductance of solenoid here: https://brainly.com/question/25484149

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