Amount of energy stored in a magnetic field

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SUMMARY

The energy stored in the magnetic field of an air-core solenoid with 57 turns, measuring 4.96 cm in length and 1.46 cm in diameter, carrying a current of 0.634 A is calculated to be 2.7355 μJ. The inductance was determined using the formula Induction = μ₀ (N²/L) A, resulting in an inductance of 1.3785e-5 H. The magnetic field strength was also calculated to be 9.09795e-4 T, although the focus remained on energy storage calculations. The provided calculations and formulas are accurate for determining the energy stored in the magnetic field.

PREREQUISITES
  • Understanding of solenoid physics and geometry
  • Familiarity with the concept of inductance
  • Knowledge of the formula for energy stored in a magnetic field
  • Basic proficiency in using SI units for physical quantities
NEXT STEPS
  • Research the derivation of the inductance formula for solenoids
  • Learn about the relationship between magnetic field strength and energy density
  • Explore the applications of inductance in electrical circuits
  • Study the effects of varying current on energy storage in magnetic fields
USEFUL FOR

Physics students, electrical engineers, and anyone interested in electromagnetism and energy storage in magnetic fields will benefit from this discussion.

rinarez7
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1. An air-core solenoid with 57 turns is 4.96 cm
long and has a diameter of 1.46 cm.
The permitivity of free space is 4×10−7 T·
m/A.
How much energy is stored in its magnetic
field when it carries a current of 0.634 A ?
Answer in units of μJ.
2.
B = mu-o (I) (N)/ L
Induction= mu-0 (N^2/l) A
U= (1/2) Induction (I ^2)

3. First I calculated Induction= 4pie-7 ( 57turns ^2/ .0496m) (pi(.0146^2))= 1.3785e-5

Then I used U = (1/2) induction (0.634 A ^2)= 2.7355μJ

I am on the wrong path? I thought of calculating the magentic field as well
using my first eqaution= 9.09795 e-4 T but I couldn't find the correct eqaution/ relationship to calculate the energy stored. Thanks in advance for any help!
 
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