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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 = NĀ²Ī¼A/l

where;

  • L is inductance
  • N is number of turns
  • A is area
  • l is average length = 2Ļ€r

NĀ²Ī¼A = LI

NĀ² = LI/Ī¼A

NĀ² = (2.3 x 10ā»Ā³ x 2Ļ€ x 0.062)/(4Ļ€ x 10ā»ā· x 2.8 x 10ā»ā“)

NĀ² = 2,546,428.6

N = āˆš2,546,428.6

N ā‰ˆ 1,596 turns

Rate of current change

L = (emf)/I

I = (emf)/L

I = (2.6)/(2.3 x 10ā»Ā³)

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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