You've decided to protect your house by placing a 5.0-m-tall i...

Question

You've decided to protect your house by placing a 5.0-m-tall iron lightning rod next to the house. The top is sharpened to a point and the bottom is in good contact with the ground. From your research, you've learned that lightning bolts can carry up to 50 kA of current and last up to 50 μs. You don't want the potential difference between the top and bottom of the lightning rod to exceed 100 V. What minimum diameter must the rod have?

Accepted Answer

Welcome back everyone. An engineer is designing a lightning rod to protect the equipment of a radio station. The lightning rod made of copper and with a length of 3 m is placed on an isolated surface at the top of the building. The upper end of the rod is taper to form a sharp point and the lower end is connected to the earth, determine the minimum radius for the rod. If the maximum allowable potential difference between the top and the bottom of the rod is 140 volts. Assume that an average lightning bolt carries a current of 40 ks and can persist for duration up to 80 microseconds for our answer choices. A is 0.22 centimeters. B is 0.86 centimeters. C is 1.2 centimeters and D is 1.6 centimeters. Now, if we're going to make sense of our problem, it would be good for us to try and draw what's going on here. Ok. So we're talking about a lightning rod that forms a sharp point at its upper end. So it would probably look something like this. Ok. So we have our rod and we know that the bottom of it is connected to the earth. So let's represent that here. Let me just insulated. Ok. Oops, don't forget our sharp point. Probably look something like this, right? And for our lightning rod, of course, is to protect the equipment for the radio station. So if we can imagine it, you know, we have our lightning. That's not good looking lightning. Uh I think this one looks better. I'm probably spending way too much time getting the lightning, right. But you, I think you guys get the idea. Now, what do we know here? Well, we know that we're trying to find the minimum radius for our lightning rod. Ok. So we're trying to find our radius. R that's what we don't know, but we already know that our rod is 3 m long. Ok. We also know that the maximum allowable potential difference between the top and the bottom of the rod is 140 volts. So let's represent that here. Ok. And we also know for the lightning that will strike the rod that and we're assuming that it's carrying a current of 40 kg amps. So that's 40 times 10 to 3 amps and can persist for duration of 80 microseconds. So the time is 80 times 10 to the negative six seconds. Now, how are we going to find the radius, given this information? Well, let's ask ourselves, we're talking about the lightning rod. We have tons of physical information here in terms of the length of the material and so on. And we want to find the radius. Do we know how current is related to the radius for uh a material? Well, we already know that current density. OK. So recall that current density G is the current per square meter for a cross sectional area in a conductor. Right now, the reason why we're talking about the area is because this is a, we assume that this is a cylindrical cylindrical lightning rod. And that means it's cross sectional area is a circle. So we should be able to find the area and the area is what will tell us what the radius will be the radius that we need. So that's what we want. But the problem is we don't know what the current density is. We know the current, right, but we don't know the current density. So again, let's ask ourselves, what do we know about current density? Well, we also know that current density equals conductivity multiplied by the electric field strength or we can also think about it as the electric field strength divided by the resistivity of a material. Now, that's helpful because we know that the lightning rod is made of copper and the resistivity of copper equals 1.7 times 10 to the negative 8 0 m. So that we know that we know, so we can use that to help us to find the current density but again, we meet into another problem, we don't know the electric field strength. So what do we know about electric field strength that can help us? Given the information, for example, do we know anything that relates electric fuel strength to voltage? Well, again, we also know that electric field strength equals a change in voltage divided by the length of a material inside a conductor. We know our change in voltage and we know the length of our material. We know that it's 3 m. So here's what we can do now that we have a value for our electric field strength or we know how to find it. We can substitute it to get a formula for our current density and substitute that to help us eventually solve for the minimum radius. So let's go ahead and do that. So first let's substitute our value for the electric field strength into our formula for the current density because no, that tells us that our current density equals a change in voltage divided by the length multiplied by rule. OK. Let me write that a little better because all of that was being divided by rule. Now that we have that we can substitute it into our formula for current density because no, this tells us well, our change in voltage divided by our length multiplied by row equals our current divided by our area, which we know that our area is pi R squared. So if we do some transposition here. That is we are solving for our are radios. Then that means pi R squared multiplied by our changing in voltage, our voltage equals our current multiplied by the length of our material multiplied by its resistivity. Now again, we can isolate R squared. That's going to be I multiplied by L multiplied by row divided by pi multiplied by our voltage. Which then tells us that if we find the skirt of both sides, our minimum radius will be the square root of I multiplied by L multiplied by row divided by pi multiplied by V nowhere somewhere because we know what all of these values are. So let's go ahead and substitute. So now that means we're going to be finding the square root. OK, gonna be finding the square root of our current, which is 40 times 10 to the third amperes multiplied by the length of the material, which is 3 m multiplied by our resistivity of copper, which is 1.7 times 10 to the negative 8 0 m. And all of that is being divided by pi multiplied by our voltage, which is 140 volts. Now, when you put that into your calculator, you'll find that you get 2.2 times 10 to the negative 3 m. But remember all our answers were in centimeters. So if we read that as centimeters, that would be 0.22 centimeters. So what does this tell us then this tells us that the minimum radius required for the rod is 0.20 0.22 centimeters. And when we look back in our answer choices, that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Electric Field and Potential Difference

Current and Resistance

Lightning Rod Design and Function

Watch next