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MHT CET · Physics · Electromagnetic Induction

The self induction (L) produced by solenoid of length ' \(l\) ' having ' \(N\) ' number of turns and cross sectional area ' \(\mathrm{A}\) ' is given by the formula ( \(\phi=\) magnetic flux, \(\mu_0,=\) permeability of vacuum)

  1. A \(\mathrm{L}=\mathrm{N} \phi\)
  2. B \(\mathrm{L}=\mu_0 \mathrm{NA} l\)
  3. C \(\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)
  4. D \(\mathrm{L}=\frac{\mu_0 \mathrm{NA}}{l}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)

Step-by-step Solution

Detailed explanation

Magnetic field inside the solenoid is, \(\mathrm{B}=\frac{\mu_0 \mathrm{NI}}{l}\) Flux inside the coil, \(\phi=\mathrm{N}(\mathrm{BA})\)
\(\therefore \quad \phi=\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}\)
Also, self inductance, \(\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}}{\mathrm{I}}\) \(\therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)