Chapter 28, Problem 42PQ

To determine

The length and the radius of the conductor.

Answer to Problem 42PQ

The length and the radius of the conductor are $108\text{m}$ and $8.35\times {10}^{-5}\text{m}$.

Explanation of Solution

Write the resistance of the conductor.

   $R=\rho \frac{l}{A}$                                                                                                         (I)

Here, $R$ is resistance of the conductor, $l$ is length of the conductor, and $A$ is area of the conductor, and $\rho$ is resistivity.

Rearrange equation (I) for $l/A$.

   $\frac{l}{A}=\frac{R}{\rho }$                                                                                                         (II)

Write the relation for the volume of the given silver cylindrical conductor.

   $V=\frac{m}{{\rho }^{\prime }}$                                                                                                         (III)

Here, $V$ is volume of the given cylindrical conductor, $m$ is mass of the cylinder, and ${\rho }^{\prime }$ is density of silver conductor.

Alternatively the volume of the cylindrical conductor is,

   $V=lA$                                                                                                         (IV)

Write the relation for the area of the conductor.

   $\begin{array}{l}A=\pi r^2\\ r=\sqrt{\frac{A}{\pi }}\end{array}$                                                                                                       (V)

Conclusion:

Substitute $78.5\text{Ω}$ for $R$, $1.586\times {10}^{-8}\text{Ω}\cdot \text{m}$ for $\rho$ to find $l/A$ in equation (II).

  $\begin{array}{c}\frac{l}{A}=\frac{78.5\text{Ω}}{1.586\times {10}^{-8}\text{Ω}\cdot \text{m}}\\ =4.95\times {10}^9{\text{m}}^{-1}\end{array}$

Substitute $0.0250\text{kg}$ for $m$ and $10490{\text{kg/m}}^{\text{3}}$ for ${\rho }^{\prime }$ to find $V$ in equation (III).

   $\begin{array}{c}V=\frac{0.0250\text{kg}}{10490{\text{kg/m}}^{\text{3}}}\\ =2.38\times {10}^{-6}{\text{m}}^{\text{3}}\end{array}$

Combine equation (II) and (IV).

   $\begin{array}{l}\frac{lA}{A^2}=\left(4.95\times {10}^9{\text{m}}^{-1}\right)\\ lA=\left(4.95\times {10}^9{\text{m}}^{-1}\right)A^2\\ 2.38\times {10}^{-6}{\text{m}}^{\text{3}}=\left(4.95\times {10}^9{\text{m}}^{-1}\right)A^2\\ A=\sqrt{\frac{2.38\times {10}^{-6}{\text{m}}^{\text{3}}}{\left(4.95\times {10}^9{\text{m}}^{-1}\right)}}\\ =2.19\times {10}^{-8}{\text{m}}^{\text{2}}\end{array}$

Length of the wire is given by

   $l=\left(4.95\times {10}^9{\text{m}}^{-1}\right)A$

Substitute $2.19\times {10}^{-8}{\text{m}}^{\text{2}}$ for $A$ to find $l$ in above expression.

   $\begin{array}{c}l=\left(4.95\times {10}^9{\text{m}}^{-1}\right)\left(2.19\times {10}^{-8}{\text{m}}^{\text{2}}\right)\\ =108\text{m}\end{array}$

Substitute $2.19\times {10}^{-8}{\text{m}}^{\text{2}}$ for $A$ in equation (V) to find $r$.

   $\begin{array}{c}r=\sqrt{\frac{2.19\times {10}^{-8}{\text{m}}^{\text{2}}}{\pi }}\\ =8.35\times {10}^{-5}\text{m}\end{array}$

Therefore, the length and the radius of the conductor are $108\text{m}$ and $8.35\times {10}^{-5}\text{m}$.