Question: (a) Find the density $\rho$of mobile charges in a piece of copper, assuming each atom contributes one free electron. [Look up the necessary physical constants.]

(b) Calculate the average electron velocity in a copper wire 1 mm in diameter, carrying a current of 1 A. [Note:This is literally a snail'space. How, then, can you carry on a long distance telephone conversation?]

(c) What is the force of attraction between two such wires, 1 em apart?

(d) If you could somehow remove the stationary positive charges, what would the electrical repulsion force be? How many times greater than the magnetic force is it?

Write the expression for charge density.

$$\begin{gathered} \mathit{\rho }\mathbf{=}\frac{\mathbf{C}\mathbf{h}\mathbf{a}\mathbf{r}\mathbf{g}\mathbf{e}}{\mathbf{V}\mathbf{o}\mathbf{l}\mathbf{u}\mathbf{m}\mathbf{e}} \\ \mathbf{=}\left(\frac{Charge}{Atom}\right)\left(\frac{Atom}{Mole}\right)\left(\frac{Mole}{Gram}\right)\left(\frac{Gram}{Volume}\right) \\ \mathbf{=}\mathit{e}\mathit{N}\left(\frac{1}{M}\right)\mathit{d} \end{gathered}$$……. (1)

Here, is the charge of electron, is the Avogadro number, is the atomic mass copper and is the density of copper.

a)

Calculate the charge density of mobile charge in piece of copper.

Substitute $1.6\times {10}^{-19}\text{C}$for charge of electron$6.0\times {10}^{23}\text{mol}$, for Avogadro number, for atomic mass of copper and for density of copper in the equation (1)

Therefore,

b)

Write the expression for diameter of the copper wire.

$$\begin{gathered} d=1\text{mm} \\ =1\text{mm}\left(\frac{\text{1}\text{m}}{{\text{10}}^{\text{3}}\text{mm}}\right) \\ ={10}^{-3}\text{m} \end{gathered}$$

Write the expression for the radius of the copper wire.

$$\begin{gathered} r=\frac{d}{2} \\ =\frac{{10}^{-3}\text{m}}{2} \\ =5\times {10}^{-4}\text{m} \end{gathered}$$

Write the expression for area of the wire.

$A=\pi r^2$ …… (2)

Substitute for in equation (2)

$$\begin{gathered} A=\pi r^2 \\ =\pi {\left(5\times {10}^{-4}\text{m}\right)}^2 \\ =7.85\times {10}^{-7}{\text{m}}^{\text{2}} \end{gathered}$$

Write the formula for current density.

localid="1657976633620" $J=\frac{I}{A}$ ……. (3)

Here, is current through the wire and is area.

Substitute 1.0 for I and for in equation (3).

Write the relation between current density and drift velocity .

$$\begin{gathered} J=\frac{I}{A} \\ =\frac{1.0A}{7.85\times {10}^{-7}{m}^2} \\ =1.2738\times {10}^{6}A/m^2 \end{gathered}$$

Rearrange the above equation,

$\upsilon =\frac{J}{\rho }$ …… (4)

Substitute for and for in equation (4)

$$\begin{gathered} \upsilon =\frac{J}{\rho } \\ =\frac{1.2738\times {10}^6\left(\frac{\text{1}{\text{m}}^{\text{2}}}{{\text{10}}^{\text{4}}{\text{cm}}^{\text{2}}}\right)}{1.4\times {10}^4} \\ =9.1\times {10}^{-3} \end{gathered}$$

Hence, the drift velocity is $9.1\times {10}^{-3}cm/s$.

c)

Calculate the separation between the two wires.

$$\begin{gathered} d=1.0\text{cm} \\ =1.0\text{cm}\left(\frac{\text{1}\text{m}}{\text{100}\text{cm}}\right) \\ =0.01\text{m} \end{gathered}$$

Write the formula for the force per unit length between two parallel wires.

$\frac{F_{mag}}{length}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}$ …… (5)

Substitute for$4\pi \times {10}^{-7}$, for current carrying two wires and for separation between the wires .

$$\begin{gathered} \frac{F_{mag}}{length}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d} \\ =\frac{\left(4\pi \times {10}^{-7}\right)}{2\pi }\frac{\left(1.0\text{A}\right)\left(1.0\text{A}\right)}{\left(1.0\text{cm}\right)} \\ =2\times {10}^{-7} \end{gathered}$$

Hence, the force of the magnetic field is $2\times {10}^{-7}$.

d)

Using Gauss Law,

Write the expression for electric field of symmetrically cylindrical change distribution.

$E=\frac{1}{2\pi {\epsilon }_0}\frac{\lambda }{d}$ …… (6)

Write the expression for the force of electric filed between the two wires.

$F_e=\frac{1}{2\pi {\epsilon }_0}\frac{{\lambda }_1{\lambda }_2}{d}$ …… (7)

Here, is the linear charge density.

Using the relation,

$\lambda =\frac{I}{\upsilon }$

Here, current is constant and is the drift velocity.

${\lambda }_1={\lambda }_2=\frac{I}{\upsilon }$

Here, .$I_1=I_2$

Then equation (7) is reduced as,

$F_e=\frac{1}{{\upsilon }^2}\left(\frac{1}{2\pi {\epsilon }_0}\right)\frac{I_1{I}_2}{d}$ …… (8)

Here,

$$\begin{gathered} c^2=\frac{1}{{\mu }_0{\epsilon }_0} \\ \frac{1}{{\epsilon }_0}={\mu }_0{c}^2\backslash \end{gathered}$$

Hence, the equation (8) is reduced as,

$F_e=\frac{1}{{\upsilon }^2}\left(\frac{{\mu }_0{c}^2}{2\pi }\right)\frac{I_1{I}_2}{d}$ …… (9)

Now, Divide equation $F_e=\frac{1}{{\upsilon }^2}\left(\frac{{\mu }_0{c}^2}{2\pi }\right)\frac{I_1{I}_2}{d}$with $F_{mag}=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}$

$$\begin{gathered} \frac{F_e}{F_{mag}}=\frac{\frac{1}{{\upsilon }^2}\left(\frac{{\mu }_0{c}^2}{2\pi }\right)\frac{I_1{I}_2}{d}}{\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}} \\ \frac{f_{ele}}{f_{mag}}=\frac{c^2}{{\upsilon }^2} \end{gathered}$$

localid="1657980326651" $$\begin{gathered} \frac{F_e}{F_{mag}}=\frac{\frac{1}{{\upsilon }^2}\left(\frac{{\mu }_0{c}^2}{2\pi }\right)\frac{I_1{I}_2}{d}}{\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}} \\ \frac{f_{ele}}{f_{mag}}=\frac{c^2}{{\upsilon }^2} \end{gathered}$$$$\begin{gathered} \frac{F_e}{F_{mag}}=\frac{\frac{1}{{\upsilon }^2}\left(\frac{{\mu }_0{c}^2}{2\pi }\right)\frac{I_1{I}_2}{d}}{\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{d}} \\ \frac{f_{ele}}{f_{mag}}=\frac{c^2}{{\upsilon }^2} \end{gathered}$$ …… (10)

Thus, the ratio of forces of electric and magnetic field is 1.1 x 10^25.

$\begin{array}{rcl}\frac{f_{ele}}{f_{mag}}& =& 1.1\times {10}^{25}\\ & =& \left(1.1\times {10}^{25}\right)f_{mag}\\ & =& \left(1.1\times {10}^{25}\right)(2\times {10}^{-7}N/cm)\\ & =& 2\times {10}^{18}N/cm\\ & & \\ & & \end{array}$

Thus, the magnitude of the electric field is 2 X 10¹⁸.