The logic of the self inductance formula

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SUMMARY

The self-inductance formula M_{12}=\frac{N_2\phi _{12}}{I_1} is derived from the relationship between magnetic flux and current in a solenoid. The magnetic field inside the solenoid is expressed as B=\frac{\mu_0 NI}{l}, where \mu_0 is the permeability of free space, N is the number of turns, I is the current, and l is the length of the solenoid. The total magnetic flux \Phi is calculated as \Phi=\frac{\mu_0 N^2 I A}{l}, where A is the cross-sectional area. The inductance is defined as the total flux divided by the current, establishing a clear logic behind the formula.

PREREQUISITES
  • Understanding of electromagnetic theory
  • Familiarity with solenoid geometry and parameters
  • Knowledge of magnetic flux and its calculations
  • Basic grasp of inductance concepts
NEXT STEPS
  • Study the derivation of the self-inductance formula in detail
  • Explore the role of permeability (\mu_0) in inductance calculations
  • Learn about the applications of inductance in electrical circuits
  • Investigate the relationship between inductance and energy storage in magnetic fields
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Students and professionals in electrical engineering, physicists studying electromagnetism, and anyone interested in understanding inductance and its applications in circuits.

nabliat
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[tex]M_{12}=\frac{N_2\phi _{12}}{I_1}[/tex]

i need to remmeber it

but i can't see the logic of
the formula
??
 
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nabliat said:
[tex]M_{12}=\frac{N_2\phi _{12}}{I_1}[/tex]

i need to remmeber it

but i can't see the logic of
the formula
??

I thought the formula looked something like: [tex]\frac{\mu_{0}N^2 A}{l}[/tex]. A is the cross section & magnetic field inside the solenoid is [tex]B=\mu_0 NI/l[/tex], flux through one loop is [tex]BA=\mu_0 N IA/l[/tex] and [tex]\Phi= \mu_0 N^2 I A/l[/tex] is the total flux. The inductance is just total flux divided by current.
 

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