Question 270: Electromagnetic Induction

Question

Inductance per unit length near the middle of a long solenoid is $$\left(\mu_0=\right.$$ permeability of free space, $$\mathrm{n}=$$ number of turns per unit length, $$\mathrm{d}=$$ the diameter of the solenoid)

Accepted Answer

The inductance long solenoid is
$$\begin{aligned}
& \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \
& \mathrm{~L}=\mu_0\left(\frac{\mathrm{N}}{l}ight)^2 	imes \pi 	imes \frac{\mathrm{d}^2}{4}
\end{aligned}$$
$$	herefore \quad$$ The inductance per unit length near the middle of a long solenoid is:
$$\frac{\mathrm{L}}{l}=\mu_0 \pi\left(\frac{\mathrm{nd}}{2}ight)^2 \quad \ldots .\left(\because \frac{\mathrm{N}}{l}=\mathrm{n}ight)$$

Answer

$$\mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)^2$$

Answer

$$4 \mu_0 \pi\left(\frac{\mathrm{nd}}{2}\right)$$

Answer

$$\left(\frac{\mu_0 \pi n d}{2}\right)$$

Answer

$$\frac{4 \mu_0 \pi}{\mathrm{n}^2 \mathrm{~d}^2}$$

Electromagnetic Induction

Inductance per unit length near the middle of a long solenoid is $\left(\mu_0=\right. permeability of free space, \mathrm{n}= number of turns per unit length, \mathrm{d}=$ the diameter of the solenoid)

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