91Ó°ÊÓ

The creation of an electromotive force across an electrical conductor in a changing magnetic field is known as electromagnetic or magnetic induction. Induction was discovered in\({\rm{1831}}\)by Michael Faraday, and it was mathematically characterized as Faraday's law of induction by James Clerk Maxwell.

Treating the lightning bolt as an infinite wire carrying some current\({\rm{I}}\). Then, the formula for the magnetic field around such wire is:

\({\rm{B}} = \frac{{{{\rm{\mu }}_{\rm{0}}}{\rm{I}}}}{{{\rm{2\pi d}}}}..............{\rm{(1)}}\)

The value of\({\rm{d}}\)is the distance from the wire.

Faraday's Law states that the voltage being induced is proportional to the rate of change of the magnetic flux, precisely

\({{\rm{E}}_{{\rm{emf}}}} = - \frac{{{\rm{d\Phi }}}}{{{\rm{dt}}}}...........{\rm{(2)}}\)

Or else, averaging over some period\({\rm{\Delta t}}\)

\({{\rm{E}}_{{\rm{emf}}}} = - \frac{{{\rm{\Delta \Phi }}}}{{{\rm{\Delta t}}}}.............{\rm{(3)}}\)

The magnetic flux\({\rm{\Phi }}\)is then given by:

\({\rm{\Phi }} = \int_{\rm{S}} {{\rm{BdS}}} .............{\rm{(4)}}\)

The value of\({\rm{S}}\)is some surface and the value of\({\rm{dS}}\)is an infinitesimal surface element proportional to the normal.

In this case, the magnetic field will be parallel to the normal at every point on the surface\({\rm{S}}\), so we can erase the vector (bold letters) notation.

\({\rm{\Phi }} = \int_{\rm{S}} {{\rm{B dS}}} .............{\rm{(5)}}\)

The current is said to be:

\(\begin{aligned}{l}{\rm{at }}{{\rm{t}}_{\rm{1}}} &= {\rm{0}}\\{{\rm{I}}_{\rm{1}}} &= {\rm{2}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}\;{\rm{A}}\end{aligned}\)

\(\begin{aligned}{l}{\rm{at }}{{\rm{t}}_{\rm{1}}} &= {\rm{25}}{\rm{.0 \mu s}}\\{{\rm{I}}_{\rm{2}}} &= {\rm{0}}\;{\rm{A}}\end{aligned}\)

With the help of the fifth equation, we can see that\({\rm{\Phi }}\)is proportional to\({\rm{B}}\).

Then, with the help of the first equation\({\rm{B}}\)is proportional to\({\rm{I}}\).

As we have\({{\rm{I}}_{\rm{2}}} = {\rm{0}}\;{\rm{A}}\).

Then, we will have:

\(\begin{aligned}{}{{\rm{B}}_{\rm{2}}}{\rm{ = 0}}\\{{\rm{\Phi }}_{\rm{2}}}{\rm{ = 0}}\end{aligned}\)

Our interval is seen straight forward, so:

\(\begin{aligned}{}{\rm{\Delta t}} &= {{\rm{t}}_{\rm{2}}} - {{\rm{t}}_{\rm{1}}}\\ &= {\rm{25 \mu s}}\end{aligned}\)

We put the values in the third equation which then yields:

\({{\rm{E}}_{{\rm{emf}}}}{\rm{ = }}\frac{{{{\rm{\Phi }}_{\rm{1}}}}}{{{{\rm{t}}_{\rm{2}}}}}...............{\rm{(6)}}\)

Now, only left \({{\rm{\Phi }}_{\rm{1}}}\) to evaluate.

(a)

Approximating the field is said to be constant on the whole loop. This is not the case in reality as we see from the first equation that the field varies with distance and the loop has a non-zero width. Then also, these kinds of approximations are very common and often quite precise.

Since the value of\({\rm{B}}\)is now constant we can pull it out of the integral with the help of the fifth equation so we have:

\({{\rm{\Phi }}_{\rm{1}}}{\rm{ = }}\int_{\rm{S}} {{{\rm{B}}_{\rm{1}}}{\rm{ dS}}} .............{\rm{(7)}}\)

The value\({\rm{S}}\)is simply said to be a loop such that the integral on the right-hand side of this equation is simple\({{\rm{r}}^{\rm{2}}}{\rm{\pi }}\).

The value of\({\rm{r}}\)is said to be the radius of the loop.

Then, plug in the values:

\(\begin{aligned}{}{\rm{\;r}} &= {\rm{1}}\;{\rm{m }}\\{\rm{d}} &= {\rm{50}}\;{\rm{m}}\end{aligned}\)

Then, using the equation for the magnetic field having the values:

\({{\rm{I}}_{\rm{1}}} = {\rm{2}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}\;{\rm{A}}\) , we get:

\({{\rm{\Phi }}_{\rm{1}}} = \left( {\frac{{{{\rm{\mu }}_{\rm{0}}}{\rm{I}}}}{{{\rm{4\pi d}}}}} \right){\rm{ \times }}{{\rm{r}}^{\rm{2}}}{\rm{\pi }}.............{\rm{(8)}}\)

Combining it with the expression for the induced voltage, we get:

\(\begin{aligned}{}{{\rm{E}}_{{\rm{ind}}}} &= \left( {\frac{{{{\rm{\mu }}_{\rm{0}}}{\rm{I}}}}{{{\rm{4\pi d\Delta t}}}}} \right){\rm{ \times }}{{\rm{r}}^{\rm{2}}}{\rm{\pi }}\\ &= \frac{{{\rm{(1}}{{\rm{0}}^{{\rm{ - 7}}\;}}{\rm{T}} \cdot {\rm{m}} \cdot {{\rm{A}}^{{\rm{ - 1}}}}{\rm{)}} \times {\rm{(2}}{\rm{.00}} \times {\rm{1}}{{\rm{0}}^{\rm{6}}}\;{\rm{A)}} \times {{{\rm{(0}}{\rm{.50}}\;{\rm{m)}}}^{\rm{2}}} \times {\rm{\pi }}}}{{{\rm{(50}}\;{\rm{m)}} \times \left( {{\rm{25}}\;{\rm{\mu s}}\left( {\frac{{{\rm{1}}{{\rm{0}}^{ - {\rm{6}}}}{\rm{s}}}}{{{\rm{1}}\;{\rm{\mu s}}}}} \right)} \right)}}\\ &= {\rm{251}}{\rm{.3}}\;{\rm{V}}\end{aligned}\)

Therefore, the voltage induced is: \({\rm{251}}{\rm{.3}}\;{\rm{V}}\).

(b)

This kind of situation produces noticeable consequences when, for example, the lightning would hit next to a house whose electrical installation is mostly in a vertical plane. Then, the lightning strike would produce an unwanted voltage in the circuits that must be having perceivable consequences and even cause damage.

Therefore, lightning strikes would produce noticeable consequences if on next to a building whose electrical installations are mostly said to be on a vertical plane.