91Ó°ÊÓ

a) Radius of the string,$r=2.00\mathrm{mm}\mathrm{or}2×{10}^{-3}\mathrm{m}$

b) Length of the string,$L=0.800\mathrm{km}\mathrm{or}800\mathrm{m}$

c) Thickness of the water layer,$t=0.500\mathrm{mm}\mathrm{or}0.5×{10}^{-3}\mathrm{m}$

d) Resistivity,$\mathrm{ÒÏ}=150\mathrm{Î\copyright }.\mathrm{m}$

d) Potential difference,$\mathrm{V}=160\mathrm{MV}\mathrm{or}160×{10}^6\mathrm{V}$

The current is the rate of flow of charges per unit of time. The current density is the rate of flow of charges per unit time per unit cross-section area.

First, we have to find the cross-sectional area of the layer of water. Then, we have to find the resistance of the wet string by using the values of resistivity, length, and the cross-sectional area. We can use Ohm’s law to find the current through the water layer.

Formulae:

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

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

The voltage formula using Ohm’s law, V = IR …(ii)

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

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

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

The changed cross-sectional area of the layer of water is given using equation (iii) as follows:

(where,r is the radius of the kite string and t is the thickness of the water layer.)

$$\begin{gathered} A=\mathrm{Ï€}{\left(r+t\right)}^2-{\mathrm{Ï€°ù}}^2 \\ =\mathrm{Ï€}{\left(2×{10}^{-3}\mathrm{m}+0.5×{10}^{-3}\right)}^2-\mathrm{Ï€}{\left(2×{10}^{-3}\mathrm{m}\right)}^2 \\ =7.07×{10}^{-6}{\mathrm{m}}^2 \end{gathered}$$

Now, theresistance of the wire is given using equation (i) as follows:

$$\begin{gathered} R=\frac{150\mathrm{Î\copyright }.\mathrm{m}×800\mathrm{m}}{7.07×{10}^{-6}{\mathrm{m}}^2} \\ =1.698×{10}^{10}\mathrm{Î\copyright } \end{gathered}$$

Finally, using the above values in equation (ii), we can get the value of the current through the layer as follows:

$$\begin{gathered} I=\frac{160×{10}^6\mathrm{V}}{1.68×{10}^{10}a} \\ =9.42×{10}^{-3}\mathrm{A} \end{gathered}$$

Hence, the value of the current is $9.42×{10}^{-3}\mathrm{A}$.