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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