# Current through swimmer due to lightning strike

• Ampere
In summary: I really appreciate it.In summary, the problem discusses a swimmer at a distance of 35m from a lightning strike to the water, with current of 78 kA. The water has a resistivity of 30 Ohm*m and the swimmer's resistance is 4 kOhm. The current spreads through the water over a hemisphere centered on the strike point. The current through the swimmer is determined to be 2.66 A. The solution presented in the problem makes several assumptions in order to simplify the problem, such as the swimmer not disturbing the hemispherical symmetry of the current and the current density being constant at a given radial distance. These assumptions may not reflect real-world situations accurately.
Ampere

## Homework Statement

Figure 26-30 shows a swimmer at distance D=35m from a lightning strike to the water, with current I=78kA. The water has resistivity 30 Ohm*m, the width of the swimmer along a radial line from the strike is 0.7m, and his resistance across that width is 4 kOhm. Assume that the current spreads through the water over a hemisphere centered on the strike point. What is the current through the swimmer?

## Homework Equations

V = IR
R = pL/A
Surface area of a hemisphere = 2*pi*r^2

## The Attempt at a Solution

Assume the swimmer has a body surface area of 'A' underwater.
Then the resistance of a hypothetical 'wire' of seawater with cross-sectional area A will be p*L/A = 30*35/A

The current through this 'wire' will be the current of the lightning strike multiplied by the fraction of the surface area of the hemisphere occupied by the swimmer: Iwire = Istrike*(A/(2*pi*L^2)) = (78,000*A)/(2*pi*35^2)

The voltage applied to the swimmer is V = IR = (78,000*A*30*35)/(2*pi*35^2*A). The area of the swimmer cancels out and V is approximately 10,640 volts.

Then the current through the swimmer is I = V/R = 10,640/4000 = 2.66 A.

Did I make a mistake by not using the radial width of the swimmer? Also, my solution implies that the current through a swimmer is proportional to 1/R, where R is the distance between the swimmer and the lightning strike. Is that the right end behavior?

#### Attachments

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I think you'd want to look for the potential difference in the water between the two radii 35 m and (35 + 0.7) m. Then take ΔV/R to find the current through the swimmer.

The answer for this question in the solution manual of Fundamentals of Physics (Halliday) is different, the current across the swimmer is 5.22x10^-2 A. I just didn't understand why the current is constant in a given distance r.

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Hi Anne Leite, Welcome to Physics Forums!

The solution presented in your figure is making some helpful but probably unrealistic simplifications in order to make the problem tractable.

The solution assumes that the presence of the swimmer in the water isn't disturbing the hemispherical symmetry of the current radiating outward from lightning strike, that the the swimmer is somehow conforming to the hemispherical shape between two fixed radii, and that he presents a single uniform resistance.

What is assumed constant for a given radial distance is the current density, J. This corresponds to the hemispherical symmetry assumption.

Hi Gneill! Why is the current density assumed to be constant at a radial distance? I didn't understand why that can be done. I know it is related to the hemispheric symmetry...

The current spreads out radially from the point of impact of the lightning strike. So the entire current is spread over the surface of a hemisphere at any given radius. Assuming that this happens symmetrically, the current density at a given radius will be a constant value over the hemispherical surface with that radius. The larger the distance, the larger hemispherical surface and the more the current is spread out (goes as the inverse square of the radius).

But why can you assume it spreads symmetrically?
Sorry for the late response.

Anne Leite said:
But why can you assume it spreads symmetrically?
Sorry for the late response.
Because the water represents a uniform medium. There's no preferred path; every direction in which the current can be conducted has the same resistivity. An additional assumption would be that it has zero potential at infinity.

## 1. How does lightning strike affect swimmers?

Lightning strike can affect swimmers by creating a path of electric current through the body, which can cause serious injury or death. The current can disrupt the body's nervous system and even stop the heart.

## 2. What factors increase the risk of being struck by lightning while swimming?

Factors that increase the risk of being struck by lightning while swimming include being in open water, being in a body of water surrounded by tall objects, and being in or near a storm. Lightning is more likely to strike in these conditions.

## 3. How does water impact the conductivity of electricity during a lightning strike?

Water is a good conductor of electricity, meaning it allows electricity to flow through it easily. This means that if lightning strikes a body of water, the electric current can travel through the water and potentially put swimmers at risk of being shocked.

## 4. What should swimmers do in the event of a lightning storm?

Swimmers should immediately get out of the water and seek shelter indoors or in a hard-top vehicle. If these options are not available, they should crouch down with their feet together and head tucked in until the storm passes. Avoid tall objects and open spaces.

## 5. Is it safe to swim during a lightning storm if the water is not directly hit?

No, it is not safe to swim during a lightning storm even if the water is not directly hit. Lightning can travel through water and put swimmers at risk. It is important to always follow proper lightning safety precautions and avoid swimming during a storm.

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