Question 273: Electromagnetic Induction

Question

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

Accepted Answer

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

Answer

$$\mathrm{L}=\mathrm{N} \phi$$

Answer

$$\mathrm{L}=\mu_0 \mathrm{NAl}$$

Answer

$$\mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}$$

Answer

$$\mathrm{L}=\frac{\mu_0 \mathrm{NA}}{l}$$

Electromagnetic Induction

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

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