3 Engineering Example 5

3.1 The current continuity equation

Introduction

When an electric current flows at a constant rate through a conductor, then the current continuity equation states that

${\oint }_S\underset{̲}{J}\cdot \underset{̲}{dS}=0$

where $\underset{̲}{J}$ is the current density (or current flow per unit area) and $S$ is a closed surface. The equation is an expression of the fact that, under these conditions, the current flow into a closed volume equals the flow out.

Problem in words

A person is standing nearby when lightning strikes the ground. Find the voltage between the feet of the person.

Figure 14:

Mathematical statement of problem

The current from the lightning dissipates radially (see Fig 14).

1. Find a relationship between the current $I$ and current density $J$ at a distance $r$ from the strike by integrating the current density over the hemisphere $I={\int \; <!--nolimits\; -->}_S\underset{̲}{J}\cdot \underset{̲}{dS}$
2. Find the field $\underset{̲}{E}$ from the equation $E=\frac{\rho I}{2{\pi }^{2}r}$ where $E=\left|\underset{̲}{E}\right|$ and $I$ is the current.
3. Find $V$ from the integral ${\int \; <!--nolimits\; -->}_{R_1}^{R_2}\underset{̲}{E}\cdot \underset{̲}{dr}$

Mathematical analysis

Imagine a hemisphere of radius $r$ level with the surface of the ground so that the point of lightning strike is at its centre. By symmetry, the pattern of current flow from the point of strike will be uniform radial lines, and the magnitude of $\underset{̲}{J}$ will be a constant, i.e. over the curved surface of the hemisphere $\underset{̲}{J}=J\underset{̲}{\stackrel{̂}{r}}$ .

Since the amount of current entering the hemisphere is $I$ , then it follows that the current leaving must be the same i.e.

since the surface area of a sphere is $4\pi r^2$ . Therefore

$J=\frac{I}{2\pi r^2}$

Note that if the current density $\underset{̲}{J}$ is uniformly radial over the curved surface, then the electric field $\underset{̲}{E}$ must be also, i.e. $\underset{̲}{E}=E\underset{̲}{\stackrel{̂}{r}}$ . Using Ohm’s law

$\underset{̲}{J}=\sigma \underset{̲}{E}\text{ or }E=\rho J$

where $\sigma$ = conductivity = $1∕\rho$ . Hence

$E=\frac{\rho I}{2\pi r^2}$

The voltage difference between two points at radii $R_1$ and $R_2$ from the lightning strike is found by integrating $\underset{̲}{E}$ between them, so that

Interpretation

Suppose the lightning strength is current $I=10,000$ A, the person is 12 m away with feet 0.35 m apart, and the resistivity of the ground is 80 $\Omega$ m. Clearly, the worst case (i.e. maximum voltage) would occur when the difference between $R_1$ and $R_2$ is greatest, i.e. $R_1$ =12 m and $R_2$ =12.35 m which would be the case if both feet were on the same radial line. The voltage produced under these circumstances is

Task!

For $\underset{̲}{F}=\left(x^2+y^2\right)\underset{̲}{i}+\left(x^2+z^2\right)\underset{̲}{j}+2xz\underset{̲}{k}$ and $S$ being the rectangle bounded by $\left(1,0,1\right)$ , $\left(1,0,-1\right)$ , $\left(-1,0,-1\right)$ and $\left(-1,0,1\right)$ find the integral ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}$

Answer

Task!

For $\underset{̲}{F}=\left(x^2+y^2\right)\underset{̲}{i}+\left(x^2+z^2\right)\underset{̲}{j}+2xz\underset{̲}{k}$ and $S$ being the rectangle bounded by $\left(1,0,1\right)$ , $\left(1,0,-1\right)$ , $\left(-1,0,-1\right)$ and $\left(-1,0,1\right)$ (i.e. the same $\underset{̲}{F}$ and $S$ as in the previous Task), find the integral ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\times \underset{̲}{dS}$

Answer

Exercises

1. Evaluate the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{\nabla }\varphi \cdot \underset{̲}{dS}$ for $\varphi =x^{2}zsiny$ and $S$ being the rectangle bounded by $\left(0,0,0\right)$ , $\left(1,0,1\right)$ , $\left(1,\pi ,1\right)$ and $\left(0,\pi ,0\right)$ .
2. Evaluate the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\times \underset{̲}{F}\right)\times \underset{̲}{dS}$ where $\underset{̲}{F}=x{e}^y\underset{̲}{i}+z{e}^y\underset{̲}{j}$ and $S$ represents the unit square $0\le x\le 1$ , $0\le y\le 1$ .
3. Using spherical polar coordinates, evaluate the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}$ where $\underset{̲}{F}=rcos\theta \underset{̲}{\stackrel{̂}{r}}$ and $S$ is the curved surface of the top half of the sphere $r=a$ .

Answer