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### Instructions and Help about Which Form 2220 Respectively

If you haven't done so yet make sure you pause the video and try to answer the question on your own first before listening on as long as we assume that the current is uniform then we can say that the current density which is symbolized by J is equal to the current divided by the cross-sectional area of the wire now the question gives us the current as 60 amps so the numerator here is very easy it's the area that we need to come up with an expression for because we were not given that directly and to do that we note that resistance is equal to a quantity known as a resistivity multiplied by the length of the wire divided by its cross-sectional area what we can do is actually solve this equation for the area a so to do that perhaps we can multiply both sides of the equation by a so that it cancels out on the right-hand side and then divide both sides of the equation by the resistance R and so we can see that the area is equivalent to this expression right here and we're going to see that that's going to be more useful to us so let's substitute that in for the area in our current density equation and we are left with what is known as a complex fraction basically a fraction within a fraction let's multiply the denominator of our complex fraction by the resistance and also the numerator by the resistance that way it will cancel out in the denominator here and so now we have this expression for the current density and we're going to see that this is indeed more useful to us because if we look carefully we have this quantity right here and that quantity would be known as the resistance per unit length and as we can see in the question that was given to us to have a value of 0.5 ohms per kilometer now it's going to be useful to actually convert the point 1 5 ohms per kilometer into ohms per meter since meters is the standard unit so we'll say that 1 kilometer is equivalent to a thousand meters kilometers will cancel out and this would equal point zero zero zero one five ohms per meter so that's going to be the value for the resistance per unit length we're going to be filling that in for this entire expression in the parentheses there the current as stated was sixty ohms excuse me 60 amps the resistivity would be a value that you would have to look up for copper and in your textbook there should be a table that looks something like the following and so we can see for copper that the resistivity is one point six nine multiplied by 10 to the negative eight and then the units are ohms times meters and then again we're multiplying by the resistance per unit length which we had just figured out was point zero zero zero one five ohms per meter now if we study this expression we can see that the ohms are going to cancel out and then in the bottom here we have meters times meters that's going to be meters squared in the numerator we have amps so basically the unit is going to come out two amps on the top and meters squared on the bottom we could then plug this expression into our calculators and we should get about five point three three times ten to the fifth amps per meter squared so this is the correct answer to Part A now on to Part B which asks for the mass per unit length which they're using the letter lambda to represent so this would be mass divided by length now we're going to be making two substitutions here one for mass one for length and for the mass we recall that density is equal to mass divided by volume and if we multiply both sides of that equation by the volume then we can see that mass is equal to density times volume so let's replace the mass with density times volume and indeed the volume is going to need to be replaced because we don't have the volume and in general let's say the wire was cylindrical e shaped which is a good assumption for most of these problems what we have here is the length of the wire and then we have a cross sectional area right here two circular cross sectional area and as long as we assume this wire is a perfect cylinder we can say that the is equal to the cross-sectional area a times the length so we're going to be replacing V with the cross-sectional area times the length so let's go ahead and do that and now we turn to the denominator of this equation and that has length in the denominator let's come back to the resistance equation that we looked at earlier so here is that equation what we want to do is solve this for length so let's multiply both sides by the area and then divide by the resistivity which is that Greek letter Rho so it'll cancel out on the right there so length would be this expression right here we're going to substitute that in for the length of our current equation we have another complex fraction going here so let's multiply the denominator by the resistivity so it cancels here and then we can see conveniently that the areas are going to cancel out and so we're left with the density times the length times the resistivity all divided by the resistance if we look at this equation carefully we can see that we have lengths per resistance now the question gave us resistance per unit length what that meant was that the resistance divided by the length was equal to point zero zero.