Example Problem - Toroid: Difference between revisions
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(New page: <math>r_m=(\frac{1}{2})\frac{OD + ID}{2}</math> <math> l_m = { 2 \pi r_m} = 0.141\ </math> <math>H_m=(\frac{Ni}{l_m})</math> <math>H_m=\frac{20 x 2.5}{.141}= 354.6\ A /m)</math>) |
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'''Problem:''' |
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Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A. |
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'''Solution:''' |
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Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the radius by |
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<math>r_m=(\frac{1}{2})\frac{OD + ID}{2}</math> |
<math>r_m=(\frac{1}{2})\frac{OD + ID}{2}</math> |
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Revision as of 19:35, 18 January 2010
Problem:
Let look at a coil around a toroid shown in the figure below. The coil has N = 20 turns around the toroid. The toroid has an inside diameter of ID = 4 cm and an outside diameter OD = 5 cm. Determine the field intensity H along the mean path length within the toroid with a current i = 2.5 A.
Solution:
Do symmetry the magnetic field intensity Hm along a circular contour within the toroid is constant. We can find the radius by