What is the magnitude of the magnetic field inside the toroid

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SUMMARY

The magnitude of the magnetic field inside a toroid with a square cross-section of 5.00 cm and an inner radius of 16.0 cm, carrying a current of 0.700 A, is calculated to be 5.25e-4 T at the inner radius. The user attempted to calculate the magnetic field at the outer radius using the formula B = PERMIVITY * n * I, where n is the number of turns per unit length. However, the user incorrectly calculated n and sought clarification on the correct approach to determine the magnetic field at the outer radius.

PREREQUISITES
  • Understanding of magnetic fields in toroidal structures
  • Familiarity with the formula B = μ₀ * n * I
  • Knowledge of the concept of permeability (PERMIVITY)
  • Basic principles of electromagnetism
NEXT STEPS
  • Calculate the magnetic field at the outer radius of the toroid using the correct value of n
  • Review the derivation of the magnetic field formula for toroids
  • Explore the effects of varying the number of turns on the magnetic field strength
  • Investigate the impact of different cross-sectional shapes on magnetic field distribution
USEFUL FOR

Physics students, electrical engineers, and anyone studying electromagnetism or designing inductive components.

mr_coffee
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A toroid having a square cross section, 5.00 cm on a side, and an inner radius of 16.0 cm has 600 turns and carries a current of 0.700 A. (It is made up of a square solenoid bent into a doughnut shape.)

(a) What is the magnitude of the magnetic field inside the toroid at the inner radius?
5.25e-4 T
(b) What is the magnitude of the magnetic field inside the toroid at the outer radius?
? T


I keep missing part B. I tried:
B = PERMIVITY*n*I
where n = N/(2PI*R);
N is the number of turns:
R is the radius of the toroid. PERMITIVITY = 4PIE-7;
So i plugged in
n = 600/(2PI*.05)
B = 4PIE-7*.700*n which was wrong! any ideas?
 
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mr_coffee said:
square cross section, 5.00 cm on a side, and an inner radius of 16.0 cm
So what is the outer radius?
 

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