Chapter 21, Problem 43P

To determine

The length of the plastic tube, length of required copper wire, and resistance of the solenoid.

Answer to Problem 43P

Solution:

46.2 m, 21.5 km, $70.14\Omega$

Explanation of Solution

Given:

Diameter of plastic tube = 12 cm

Self-inductance of solenoid 1.0 H

Diameter of copper wire 0.81 mm

Formula used:

$\begin{array}{l}\text{Self inductance of an air filled solenoid}\\ L=\frac{{\mu }_o{N}^{2}A}{l}\end{array}$

$\begin{array}{l}\text{The length of the wire is equal to the number of turns }\times \text{ circumference of the turn}\text{.}\\ L_{plastic}=\frac{l}{d_{copper}}\pi D_{plastic}\end{array}$

$\begin{array}{l}\text{Resistance of solenoid is}\\ R=\frac{\rho l}{A}\end{array}$

Calculation:

$\begin{array}{l}\text{Self inductance of an air filled solenoid}\\ L=\frac{{\mu }_o{N}^{2}A}{l}={\mu }_o{n}^{2}Al=\frac{{\mu }_{o}Al}{d^2}\\ l=\frac{L{d}^2}{{\mu }_o\pi r^2}\\ =\frac{\left(1\text{H}\right){(0.81\times {10}^{-3}\text{m)}}^2}{\left(4\pi \times {10}^{-7}\text{T}\text{.m/A}\right)\pi {\left(6\times {10}^{-2}\text{m}\right)}^2}=46.21\text{m}\\ \text{The length of the wire is equal to the number of turns }\times \text{ circumference of the turn}\text{.}\\ L_{plastic}=\frac{l}{d_{copper}}\pi D_{plastic}\\ =\frac{\text{46}\text{.21m}}{0.81\times {10}^{-3}\text{m}}\pi (0.12\text{m)=21}\text{.5 km}\\ \text{Resistance of solenoid is}\\ R=\frac{\rho l}{A}=\frac{\rho l}{\pi \left(d^2{}_{copper}/2\right)}=\frac{\left(1.68\times {10}^{-8}\Omega \text{m}\right)\left(21.5\text{km}\right)}{\pi {\left(0.405\times {10}^{-3}\text{m}\right)}^2}=70.14\Omega \end{array}$