Find field intensity at center of toroid?

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SUMMARY

The discussion centers on calculating the magnetic field intensity (H) at the inner diameter of a toroid with an inner diameter of 3.1 cm and an outer diameter of 4 cm, carrying 8 amps through 41 turns. The correct approach involves using the inner radius (Ri) instead of the mean radius (Rm) for accurate results. The final calculation yields H = 3367.92 A/m when using Ri, correcting the initial miscalculation of 367.62 A/m. The importance of using the correct radius is emphasized for precise magnetic field intensity determination.

PREREQUISITES
  • Understanding of Ampere's Law
  • Familiarity with toroidal geometry
  • Knowledge of magnetic field intensity calculations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the derivation of Ampere's Law in relation to magnetic fields
  • Learn about the differences between mean radius and inner radius in toroidal calculations
  • Explore the impact of geometry on magnetic field intensity
  • Practice additional problems involving toroidal inductors and field calculations
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Students in physics or electrical engineering, educators teaching electromagnetism, and anyone involved in designing or analyzing toroidal inductors.

D.B0004
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Homework Statement


I really think my answer is correct - could someone verify this or let me know where I went wrong? This is problem submitted online, the "correct" answer looks like my teacher accidentally hit the button 3 twice.
PROBLEM:
The toriod in example 5-1 from the textbook has an inner diameter of 3.1 cm and an outer diameter of 4 cm. 8 amps are flowing through 41 turns on the toroid.

What is the field intensity very close to the inside diameter (A/m)?

See attachment for figure.

Homework Equations


Rm = 0.5*(Di+Do)/2
H = Ni/L
L = 2*PI*Rm

Simply find H.

The Attempt at a Solution


I plugged in the numbers many times and keep getting 367.62. My teacher got 3367.92. Looks like the 3 was clicked twice? Any help appreciated.
 

Attachments

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I'm getting the same answer as your teacher. Note that you are asked to find the field "very close to the inside diameter".
 
Okay. Here are my steps
1) Find Rm = (0.01775)/4= 0.01775
2) Find H = Ni*I/(2*PI*Rm) = 42*8/(2IPI*.01775 = 3012.8

I forgot to put the I in my first answer.

I don't see the importance of the "very close to the inside diameter" part and how that effects the equation.
 
D.B0004 said:
Okay. Here are my steps
1) Find Rm = (0.01775)/4= 0.01775
2) Find H = Ni*I/(2*PI*Rm) = 42*8/(2IPI*.01775 = 3012.8

You are using the mean value of R: Rm. But you need to use the value of R that corresponds to being close to the inner diameter.

Did you derive the formula for H in class using Ampere's law, or were you just given the formula without a derivation?
 
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We were not given it. Maybe they just assume I could derive it? Not sure. But I think I see what you are saying. So rather than going to the Rm, I should just use Ri -->
Ri = .031/2
Didn't derive it but that gives the right answer. Thanks.
 

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