Find field intensity at center of toroid?

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Homework Help Overview

The discussion revolves around calculating the magnetic field intensity at the center of a toroid, specifically focusing on the parameters of the toroid including its inner and outer diameters, the current flowing through it, and the number of turns. The original poster expresses uncertainty about their calculations compared to a provided answer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of the mean radius and the application of Ampere's law. There are questions about the significance of the phrase "very close to the inside diameter" and its impact on the calculations. Some participants share their steps and results, while others suggest reconsidering the radius used in the calculations.

Discussion Status

Multiple interpretations of the problem are being explored, particularly regarding the correct radius to use for calculating the field intensity. Some participants have provided steps and reasoning, while others have questioned assumptions made in the calculations. There is no explicit consensus yet, but guidance has been offered regarding the importance of using the appropriate radius.

Contextual Notes

Participants note a potential discrepancy in the provided answer and express uncertainty about the derivation of formulas used. There is also mention of homework constraints regarding the expectations for derivation versus application of formulas.

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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  • Figure from Ex 5.1.png
    Figure from Ex 5.1.png
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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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