Q36P - Fundamentals Of Physics

Question

Swimming during a storm. Figure shows a swimmer at distance D = 35.0m from a lightning strike to the water, with current l = 78 kA. The water has resistivity $\mathbf{30}\mathbf{\Omega m}$ , the width of the swimmer along a radial line from the strike is 0.70m , and his resistance across that width is $\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{k\Omega }$ . Assume that the current spreads through the water over a hemisphere centered on the strike point. What is the current through the swimmer?

Step-by-Step Solution

Answer

The current through the swimmer is ${\text{5.22×10}}^{-2}\mathrm{A}$ .

Step 1: Identification of given data

1. The distance between the swimmer and lightning strike is D = 35.0m 
2. The current produced by lightning strike is $l=78\mathrm{kA} \mathrm{or} 78\times {10}^3\mathrm{A}$ 
3. The resistivity of the water is $\mathrm{p}=30\mathrm{\Omega }.\mathrm{m}$ 
4. The width of the swimmer is $∆\mathrm{r}=0.70\mathrm{m}$ 
5. The resistance of the swimmer is $R=4.00\mathrm{k\Omega } \mathrm{or} 4.00\times {10}^3\mathrm{\Omega }$  

Step 2: Significance of current density and Ohm’s law

Current density is the rate of flow of charges per unit time per unit area. Ohm’s law gives the relationship between current, voltage, and resistance. The current is directly proportional to the voltage. The opposition to the flow of current is equal to the resistance. The resistance is the ratio of voltage to the current. We can use the concept of current density in terms of electric field and current. By using Ohm’s law, we can find the current through the swimmer.

 

Formulae:

The current density due to the current flow through the wire, $J=\frac{i}{A}$                           …(i)

Here, i is current, J is current density, and A is area of cross-section.

The current density due to the electric field, $J=\frac{E}{\mathrm{p}}$                                                     …(ii)

Here, J is the current density, E is the electric field, and p is resistivity.

The potential difference due to the differential equation, $V=-{\int }_D^{D+∆r}Edr$                 …(iii)

Here, E is the electric field, V is voltage, dr is small distance.

The voltage equation using Ohm’s law, V = lR                                                          …(iv)

Here, V is the potential difference, l is current, R is resistance.

The cross-sectional area of the wire, $A={\mathrm{\pi r}}^2$                                                             …(v)

Here, A is area of cross-section, r is radius

Step 3: Determining the current through the swimmer

The current spreads uniformly over the hemisphere. 

Thus, the equation of the current density through the total area can be given using equation (v) in equation (i) as follows:

 $J=\frac{i}{2{\mathrm{\pi r}}^2}$

The resistivity of water; $p_w=30\Omega m$

Now, the equation of the electric field using this above value substituted in equation (ii) can be given as follows:

$E=\left(\frac{i}{2{\mathrm{\pi r}}^2}\right)p_w$ 

Now,there exists a potential difference between the points D and $D+∆r$ .

Thus, the potential difference between the two ends of the cone is given using the above value in equation (iii) as follows:

 $$\begin{gathered} ∆V={\int }_D^{D+∆r}{p}_{w}dr \\ =-\frac{i{p}_w}{2\mathrm{\pi }}{\int }_D^{D+∆r}\frac{1}{r^2}dr \\ =-\frac{i{p}_w}{2\mathrm{\pi }}{\left(\frac{1}{r}\right)}_D^{D+∆r} \\ =-\frac{i{p}_w}{2\mathrm{\pi }}\left(\frac{1}{D+∆r}-\frac{1}{D}\right) \\ =-\frac{i{p}_w}{2\mathrm{\pi }}\left(\frac{D-D-∆r}{D(D+∆r)}\right) \\ =-\frac{i{p}_w}{2\mathrm{\pi }}\left(\frac{-∆r}{D(D+∆r)}\right) \end{gathered}$$

Now, the current value flowing through the swimmer can be given using the above value and the given data in equation (iv) as follows:

$$\begin{gathered} \mathrm{i}=\frac{{\displaystyle \frac{{\mathrm{ip}}_{\mathrm{w}}}{2\mathrm{\pi }}}\left({\displaystyle \frac{∆\mathrm{r}}{\mathrm{D}(\mathrm{D}+∆\mathrm{r})}}\right)}{\mathrm{R}} \\ =\frac{{\mathrm{ip}}_{\mathrm{w}}∆\mathrm{r}}{2\mathrm{\pi }\left[\mathrm{D}(\mathrm{D}+∆\mathrm{r})\right]\mathrm{R}} \\ =\frac{(78\times {10}^3)\mathrm{A}\times 30\mathrm{\Omega }.\mathrm{m}\times 0.70\mathrm{m}}{2\times 3.14\times \left[35.0\mathrm{m}+0.70\mathrm{m})\right]\times \left(4.00\times {10}^3\right)\mathrm{\Omega }} \\ =5.22\times {10}^{-2}\mathrm{A} \end{gathered}$$ 

Hence, the value of the current is $5.22\times {10}^{-2}\mathrm{A}$ .