Chapter 20, Problem 51P

To determine

To identify: Whether the copper wire should be long and thin, short or fat or something else.

Answer to Problem 51P

The copper wire must be short and fat with a small radius of the solenoid.

Explanation of Solution

Given:

  $1\text{ kg}$ mass of Copper is required to make solenoid producing greatest possible magnetic field.

Formula used:

The magnetic field is calculated as

  $\text{B=}\frac{{\mu }_0\text{I}}{2\pi \text{r}}$

Ohm’s law is given as $\text{V=IR}$

Calculation:

Since mass and density of Copper is fixed. The only thing variable here is its volume that is

  ${\text{V}}_{\text{Cu}}={\text{m}}_{\text{Cu}}{\rho }_{\text{Cu}}={\text{L}}_{\text{Cu}}{\text{A}}_{\text{Cu}}$

Also, voltage ${\text{V}}_0$is fixed. Resistance in the wire is calculated as

  ${\text{R}}_{\text{Cu}}={\rho }_{\text{Cu}}\frac{{\text{L}}_{\text{Cu}}}{{\text{A}}_{\text{Cu}}}$

The total length of the wire is the product of the number of turns and the circumference of the solenoid.

  $\begin{array}{l}{\text{L}}_{\text{Cu}}={\text{NC}}_{\text{S}}=\text{N}\times \text{2}\pi {\text{r}}_{\text{S}}\\ \text{N=}\frac{{\text{L}}_{\text{Cu}}}{\text{2}\pi {\text{r}}_{\text{S}}}\end{array}$

Now, using all these values in the formula to calculate the magnetic field is

  $\begin{array}{l}\text{B=}\frac{{\mu }_0{\text{I}}_{\text{Cu}}\text{N}}{{\text{L}}_{\text{S}}}={\mu }_0{\text{V}}_0\frac{\text{N}}{{\text{R}}_{\text{Cu}}{\text{L}}_{\text{S}}}=\frac{{\mu }_0{\text{V}}_0}{{\rho }_{\text{RCu}}}\frac{\text{N}}{\frac{{\text{L}}_{\text{Cu}}}{{\text{A}}_{\text{Cu}}}{\text{L}}_{\text{S}}}\\ \text{B=}\frac{{\mu }_0{\text{V}}_0}{{\rho }_{\text{RCu}}}\frac{\text{N}}{{\text{L}}_{\text{S}}}\frac{{\text{A}}_{\text{Cu}}}{{\text{L}}_{\text{Cu}}}=\frac{{\mu }_0{\text{V}}_0}{{\rho }_{\text{RCu}}}\frac{\text{N}}{{\text{L}}_{\text{S}}}\frac{{\text{m}}_{\text{Cu}}{\rho }_{\text{Cu}}}{{\text{L}}_{\text{Cu}}^2}\\ \text{B=}\frac{{\mu }_0{\text{V}}_0{\text{m}}_{\text{Cu}}{\rho }_{\text{Cu}}}{{\rho }_{\text{RCu}}}\frac{\frac{{\text{L}}_{\text{Cu}}}{2\pi {\text{r}}_{\text{S}}}}{{\text{L}}_{\text{S}}{\text{L}}_{\text{Cu}}^2}=\frac{{\mu }_0{\text{V}}_0{\text{m}}_{\text{Cu}}{\rho }_{\text{Cu}}}{2\pi {\rho }_{\text{RCu}}}\frac{1}{{\text{L}}_{\text{S}}{\text{r}}_{\text{S}}{\text{L}}_{\text{Cu}}}\end{array}$

Since, first fraction is made of constants, thus, magnetic field is proportional to $\text{B}\propto \frac{1}{{\text{L}}_{\text{S}}{\text{r}}_{\text{S}}{\text{L}}_{\text{Cu}}}$

Conclusion:

This relation suggests that the magnetic field is greatest when the radius of the solenoid is small and copper wire must be made short and fat.