91Ó°ÊÓ

1. Potential difference,$V=3\mathrm{nV}\mathrm{or}3×{10}^{-9}\mathrm{V}$
2. Length of the copper wire,$L=2.00\mathrm{cm}\mathrm{or}0.02\mathrm{m}$
3. Radius of the wire,$r=2\mathrm{mm}\mathrm{or}0.02\mathrm{m}$
4. The time of the drift,$∆t=3\mathrm{ms}\mathrm{or}3×{10}^{-3}\mathrm{s}$

The current is equal to the rate of flow of charges. The opposition to the current is resistance. The resistance is equal to the potential difference divided by the current in the conductor.

We can calculate the resistance of the copper wire by substituting the given values in the formula for resistance. Using the calculated resistance and the given potential difference in Ohm’s law, we can find the current in the wire. Using the current, we can find the charge drifting through the cross-section in 3ms.

Formulae:

The voltage equation from Ohm’s law,$V=IR$ …(¾±)

Here, $V$is a voltage, $I$is current, and $R$is resistance.

The resistance of the material, $R=\frac{L\mathrm{ÒÏ}}{A}$ …(¾±¾±)

Here, $R$is resistance, $L$is the length of the wire, $\mathrm{ÒÏ}$is the resistivity, $A$is area of cross-section.

The amount of charge flowing in the given time, $∆Q=l∆t$ …(¾±¾±¾±)

Here,$∆Q$ is the amount of charge flowing in the time ,$∆t,I$ is current.

The cross-sectional area of the wire, $A=\mathrm{Ï€}{r}^2$ …(¾±±¹)

Here, $A$is the cross-section area and$r$ is the radius.

The resistivity of the copper from the table $26-1,\mathrm{ÒÏ}=1.69×{10}^{-8}\mathrm{Î\copyright }·\mathrm{m}$

Substituting the value of area from equation (iv) and the given values in equation (ii), we can get the value of the resistance of the copper wire as follows:

$$\begin{gathered} R=\frac{0.02\mathrm{m}×1.69×{10}^{-8}\mathrm{Î\copyright }·\mathrm{m}}{\mathrm{Ï€}×{\left(0.002\mathrm{m}\right)}^2} \\ =2.69×{10}^{-8}\mathrm{Î\copyright } \end{gathered}$$

Now, substituting the value of resistance in equation (i), we can get the drift current value as follows:

$$\begin{gathered} I=\frac{3×{10}^{-9}\mathrm{V}}{2.69×{10}^{-5}\mathrm{Î\copyright }} \\ =1.115×{10}^{-4}\mathrm{A} \end{gathered}$$

Now, the amount of charge drifting through the cross-section is given using equation (iii) as follows:

$$\begin{gathered} ∆Q=1.115×{10}^{-4}\mathrm{A}×3.00×{10}^{-3}\mathrm{s} \\ =3.35×{10}^{-7}\mathrm{C} \end{gathered}$$

Hence, the value of the charge is $3.35×{10}^{-7}\mathrm{C}$.