Q22P - Fundamentals Of Physics

Question

Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose a kite string of radius 2.00 mm extends directly upward by 0.800 km and is coated with a 0.500 mm layer of water having resistivity $\mathbf{150}\mathbf{\Omega m}$. If the potential difference between the two ends of the string is 160 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).

Step-by-Step Solution

Answer

The current through the water layer is $9.42\times {10}^{-3 }\mathrm{A}$

Step 1: The given data

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

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

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

d)   Resistivity, $\mathrm{\rho }=150 \mathrm{\Omega }.\mathrm{m}$

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

Step 2: Understanding the concept of the flow of current and its density

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, $R=\frac{\mathrm{\rho }L}{A}$                                                                 …(i)

Here,$\mathrm{\rho }$ 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, $A\pi r^r$                                                              …(iii)

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

Step 3: Calculation of the current through the water layer

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.)

$$\begin{gathered} A=\mathrm{\pi }{\left(r+t\right)}^2-{\mathrm{\pi r}}^2 \\ =\mathrm{\pi }{\left(2\times {10}^{-3} \mathrm{m}+0.5\times {10}^{-3}\right)}^2-\mathrm{\pi }{\left(2\times {10}^{-3}\mathrm{m}\right)}^2 \\ =7.07\times {10}^{-6} {\mathrm{m}}^2 \end{gathered}$$

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

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

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

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

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