Question

The inductance L of a solenoid of length l , whose windings are. made of material of density D and resistivity $\rho$ is (the winding resistance is R)

• A. $\frac{\mu_0}{4\pi l}\frac{Rm}{\rho D}$
• B. $\frac{\mu_0}{4\pi r}\frac{lm}{\rho D}$
• C. $\frac{\mu_0}{4\pi l}\frac{R^2m}{\rho D}$
• D. $\frac{\mu_0}{2\pi R}\frac{lm}{\rho D}$

Answer 1

Answer: (A)

$\frac{\mu_0}{4\pi l}\frac{Rm}{\rho D}$

Answer 2

For a solenoid, $L = \mu_0 N^2\frac{A}{l}$ . If $x$ is the length of the wire and $a$ is the area of cross-section, then $R = \frac{\pi x}{a}$ and $m = axD$ $Rm = \frac{\phi x}{a} axD,x = \sqrt{\frac{Rm}{\phi D}}$ Also, $x = 2 \pi rN , N = \frac{x}{2 \pi r} \left( \therefore \, L = \frac{\mu_0N^2 A}{l} \right)$ $\therefore$ $L = \mu_0 \left( \frac{x}{2 \pi r} \right)^2 \frac{ \pi r^2}{l} = \frac{\mu_0}{4 \pi l} \frac{Rm}{\phi D}$