Imagine that a steady current I flows in a straight cylindrica...

Question

Imagine that a steady current I flows in a straight cylindrical wire of radius R₀ and resistivity ρ.

(a) If the current is then changed at a rate dI/dt, show that a displacement current ID exists in the wire of magnitude ε₀ρ(dI/dt).

(b) If the current in a copper wire is changed at the rate of 1.0 A/ms, determine the magnitude of ID.

(c) Determine the magnitude of the magnetic field BD created by ID at the surface of a copper wire with R₀ = 1.00 mm. Compare (as a ratio) BD with the field created at the surface of the wire by a steady current of 1.0 A.

Accepted Answer

Welcome back, everyone in this problem. An unknown long straight cylindrical metallic wire is connected to a three volt battery. The wire has a radius of three millimeters and a length of 4 m using a tesla meer. The magnetic field at a distance of four millimeters from the surface of the wire is measured to be 34 mill teslas. We want to determine the resistivity of the wire for our answer choices. A is 1.8 times 10 to the negative eight B is 2.6 times 10 to the negative eight C is seven times 10 to the negative eight and D is one times 10 to the negative seven. And all of these are written in 0 m. Now, let's try to visualize what's going on here to get a better idea. So we have a long straight cylindrical metallic wire that's connected to a three volt battery. So I'm going to draw the battery here. Ok. Well, I'm gonna try to draw the battery right? And if this is say our battery, then if we have our long straight wire connected to it, then it would look something like this. Mhm Yeah, something like that and let me draw another layer onto it just to show that it's gonna have a cross section and everything. Ok. Now, what do we know about our wire? Well, so far as we said, we know that the battery starting with a battery, it is a three volt battery, our wire has a radius of three millimeters. So let me put that in here. So that's three millimeters and it has a length of 4 m. So already stating what we know, we know that the radius is three times 10 to the negative 3 m for our wire. And we also know that its length is 4 m. OK? Now, we also know that the voltage of the battery is three volts. Now we want to find the resistivity of the wire. In other words, we want to find the value of rule. OK? But now let's think about it. What do we know here? What do we know about the resistivity of a material? Well, recall, OK, that the resistance is directly proportional to the resistivity of our material. And as given by the formula R equals row multiplied by L divided by A where L is the length of our material and A is its cross sectional area. So that means if we were to solve for role that is the re resistivity, then our resistivity would be equal to R multiplied by A divided by L but A is the cross sectional area. Of our wire. Do we know what it is? Well, because it's a cylindrical wire, then we know that A will be equal to pi R squared. So row is actually equal to R multiplied by pi R squared divided by L. When we think about what we have here, we know the radius, right. Uh We know the length of our wire, but we don't know the resistance of our wire. So I should rather put a question sign here. So if we can figure out the resistance we can solve to find the resistivity. Now, what do we know about resistance? Well, by S law, again, recall that resistance equals the change in the potential difference divided by the current where I is the current flowing. OK. And we know that our voltage for our battery is three volts. We know that one. But again, no, we don't know what the current is. We don't know what I is. So if we can figure out I, then we should be able to solve. Let's look on the information we already have. What else do we know? Well, we know that there's a magnetic field at a distance of four millimeters from the surface of the wire and it's measured to be 34 milli teslas. So what does that mean? If you can imagine here, let's say I were to come down here or let me draw the error a little better. Let's say I were to come around here and zoom in on what's really going on. It's as if we have our wire, OK. This is a cross section of our wire which is three millimeters and the magnetic field is at a distance of four millimeters from the surface of the wire. So if we imagine this dotted line, let me draw it a bit more ace in ji. If we imagine this to be the magnetic field, then it is four millimeters from the surface of the wire. In other words, from the surface to the outer edge would have been four millimeters. OK. Now, is there anything we know that relates the magnetic field or the strength of the magnetic field to the current that's flowing? Well, recall that the magnetic field is directly proportional to the current. In other words, if we double the current, we double the strength of the magnetic field and it's related by the formula B which is the strength of the magnetic field equals mu not multiplied by I divided by two pi Y OK. Where mu knot is the permittivity in free space, which is 1.26. OK. Or the permeability of free space? My apologies 1.26 multiplied by 10 to the negative six tesla meters per amper I of course, is the current flowing. OK. And which we don't know as we said earlier and why would be equal to Y would be equal to the distance from the center of the wire. So if our wire is three millimeters, its cross sectional, its radius is three millimeters. And from the surface to the magnetic field is four millimeters, then the distance from the center of the wire to the magnetic field would be seven millimeters. So that's seven times 10 to the negative 3 m. So when we think about it, we already know the permeability of free space, we know the distance from the center of the wire and we know the strength of the magnetic field. Let me write that down here. We know that B equals 34 milli teslas or 34 times 10 to the negative three uh teslas. So because we know those, that means we can solve for I, OK, we can solve our, I put it into our equation for R and then we can use that to help us to solve for R. So let's go ahead and do that. We have some transposition here to do. So if B equals mu not multiplied by I divided by two pi Y, then this implies that our current I equals two pi Y multiplied by B divided by mu not. So now we have a value for I check next, let's substitute that in our formula for resistance. So now this tells us that our resistance would be equal to our potential difference. Our voltage V change in vol V divided by IV divided by two pi yb divided by mu knot, which we know that we're going to reciprocate our value for I. So know that changing V multiplied by Muno divided by two pi yb. So now we have a value for R with all the known variables. So to solve for the resistivity rule, let's substitute. So no, that means rule status, then row resistivity equals R which is a change in voltage multiplied by mu knot divided by two pi YP multiplied by pi R squared. OK. And all of it was divided by L. So we can write it right here. No, we can simplify a bit further because both terms contain the constant pi so we can write our resistivity as we multiply it by mu knot R squared divided by two YBL. We're almost there because now we have all the values that we need, we know all of these values. So now we can go ahead and substitute the values to solve for the resistivity. So therefore, that means our resistivity equals our change in voltage three volts multiplied by Mu knot which we said was 1.26 times 10 to the 10 to the negative six tesla meter per pi OK. Multiplied by the radius of our wire. So that's three millimeters three times 10 to the negative 3 m. And that is going to be squared. And all of that is being divided OK by two multiplied by Y which is seven millimeters. So that's seven times 10 to the negative 3 m multiplied by our magnetic field strength which is 34 mi mi teslas. Sorry. So that's 34 times 10 to the negative three teslas multiplied by our length, the length of our wire which is 4 m. So now we've substituted all these values, we can go ahead and solve our our resistivity by putting this into our calculator. Now, when you put this into your calculator, you'll find that you get 1.8 times 10 to the negative 8 0 m. So what this means is that the resistivity of our metallic wire is 1.8 times 10 to the negative 8 0 m. When we look back on our answer choices, that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Displacement Current

Resistivity

Magnetic Field due to Current

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