Question

Find the inductance of a solenoid of length $l_0$, made of Cu windings of mass $m$. The winding resistance is equal to $R$. The diameter of solenoid << $l$. ${\rho }_0$ is resistivity of Cu and $\rho$ is density of the Cu.

• A. $\frac{{\mu }_{0}Rm}{2\pi l_0\rho {\rho }_0}$
• B. $\frac{{\mu }_{0}Rm}{4\pi l_0\rho {\rho }_0}$
• C. $\frac{{\mu }_{0}Rm}{3\pi l_0\rho {\rho }_0}$
• D. $\frac{2{\mu }_{0}Rm}{3\pi l_0\rho {\rho }_0}$

Answer 1

Answer: (B)

$\frac{{\mu }_{0}Rm}{4\pi l_0\rho {\rho }_0}$

$R=\frac{{\rho }_{0}l}{A}$
Length of the wire $l=\frac{RA}{{\rho }_0}=\frac{Rm}{{\rho }_0\rho l}$
or
$l=\sqrt{\frac{Rm}{\rho {\rho }_0}}$ where ${\rho }_0$ is
resistivity of the Cu and $\rho$ is density of the $Cul=n{l}_{0}2\pi r$ and $L={\mu }_0{n}^2{l}_0\pi r^2$ where $l_0$ is length of
the solenoid and r is the radius of the solenoid. Then $l=l_{0}2\pi r\sqrt{\frac{L}{{\mu }_0{l}_0\pi r^2}}$

or $l=\sqrt{\frac{4\pi L{l}_0}{{\mu }_0}}$

or $l=\sqrt{\frac{Rm}{\rho {\rho }_0}}$

or $L=\frac{{\pi }_{0}Rm}{4\pi l_0\rho {\rho }_0}$