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A toroird 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 calulate the magnetic field inside the toroid at (a) the inner and (b) the outer radius of the toroid

Hamza Khan , 9 Years ago
Grade 12th pass
anser 2 Answers
Vikas TU
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
 
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