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A 2.30 mH toroidal solenoid has an average radius of 6.20 cm and a cross-sectional area of 2.80 cm2.
a) How many coils does it have? In calculating the flux, assume that B is uniform across a cross section, neglect the variation of B with distance from the toroidal axis.
b) At what rate must the current through it change so that a potential difference of 2.60 V is developed across its ends?

Respuesta :

onyebuchinnaji

(a) The number of turns of the coil is determined as 1,596 turns.

(b) The rate of change of current is determined as 1,130.43 A/s.

Number of turns of the solenoid

L = N²μA/l

where;

  • L is inductance
  • N is number of turns
  • A is area
  • l is average length = 2πr

N²μA = LI

N² = LI/μA

N² = (2.3 x 10⁻³ x 2π x 0.062)/(4π x 10⁻⁷ x 2.8 x 10⁻⁴)

N² = 2,546,428.6

N = √2,546,428.6

N ≈ 1,596 turns

Rate of current change

L = (emf)/I

I = (emf)/L

I = (2.6)/(2.3 x 10⁻³)

I = 1,130.43 A/s

Thus, the number of turns of the coil is determined as 1,596 turns.

The rate of change of current is determined as 1,130.43 A/s.

Learn more about number of turns here: https://brainly.com/question/25886292

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