Magnetic Field inside a Solenoid (Please explain this answer to me)

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SUMMARY

The magnetic field inside a solenoid can be calculated using the formula B = μ₀ * I * N, where B is the magnetic field, μ₀ is the permeability constant (1.257E-6 T·m/A), I is the current (2.0 A), and N is the number of turns (500). For a solenoid measuring 3.0 cm in length and 0.50 cm in radius, the resulting magnetic field is 4.2 x 10-2 T. The magnetic field is independent of the solenoid's cross-sectional area and is applicable to ideal solenoids.

PREREQUISITES
  • Understanding of electromagnetic theory
  • Familiarity with solenoid geometry
  • Knowledge of the permeability constant
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the magnetic field formula for solenoids
  • Explore the concept of magnetic fields in different geometries
  • Learn about the applications of solenoids in electromagnetism
  • Investigate the effects of varying current and number of turns on magnetic field strength
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Students studying electromagnetism, physics educators, and anyone interested in the practical applications of solenoids in electrical engineering.

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Homework Statement


Q5. A solenoid is 3.0 cm long and has a radius of 0.50 cm. It is wrapped with 500 turns of wire carrying a current of 2.0 A. T


Homework Equations



Magnetic field of an ideal solenoid = permeability constant (1.257E-6) * current * number of turns

The Attempt at a Solution



The given answer is: 4.2 x 10-2 T

I don't understand how to get this answer. From what my book says, the magnetic field is independent of cross-sectional area and only deals with semi-infinite solenoids. Please help!
 
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err nvm, I figured it out. I was seriously staring at this thing for two hours and then finally figure it out after I post it ><
 

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