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)
- A \(\mathrm{L}=\mathrm{N} \phi\)
- B \(\mathrm{L}=\mu_0 \mathrm{NA} l\)
- C \(\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)
- D \(\mathrm{L}=\frac{\mu_0 \mathrm{NA}}{l}\)
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, \(\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}}{\mathrm{I}}\) \(\therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)
\(\therefore \quad \phi=\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}\)
Also, self inductance, \(\mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{\frac{\mu_0 \mathrm{~N}^2 \mathrm{IA}}{l}}{\mathrm{I}}\) \(\therefore \quad \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}\)
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