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A toroid has a 4.63 cm square cross section, an inside radius of 15.0 cm, 527 turns of wire, and a current of 0.725 A. What is the magnetic flux through the cross section

Respuesta :

Using the equation, B = (μoIN/2πr)

The inner radius is r = 16.2 cm,

so the field there is

B = (4π 10-7).(0.813).(535)/2π(0.162)

= 5.37 × 10-4 T

The outer radius is r = 16.2 + 5.2 = 21.4 cm.

The field there is B = (4π 10-7).(0.813).(535)/2π(0.214)

= 4.06 × 10-4 T.

A toroid having a square cross-section, 5.20cm on edge and an inner radius of 16.2cm has 535 turns and carries a current of 813mA to calculate the magnetic.

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