Find field intensity at center of toroid?

AI Thread Summary
The discussion revolves around calculating the magnetic field intensity at the center of a toroid with specific dimensions and current. The original poster believes their answer of 367.62 A/m is correct, while their teacher's answer is significantly higher at 3367.92 A/m. It is clarified that the correct approach requires using the radius corresponding to the inner diameter rather than the mean radius. By applying the correct radius, the user finds the accurate field intensity. The importance of deriving the formula using Ampere's law is also highlighted, indicating a gap in understanding the underlying principles.
D.B0004
Messages
16
Reaction score
0

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

  • Problem Statement.png
    Problem Statement.png
    39.8 KB · Views: 472
  • Figure from Ex 5.1.png
    Figure from Ex 5.1.png
    37.4 KB · Views: 509
Physics news on Phys.org
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?
 
  • Like
Likes D.B0004
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top