A 1.50-m cylinder of radius 1.10 cm is made of a complicated...

A 1.50-m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resistivity depends on the distance $x$ from the left end and obeys the formula $\rho (x) = a + bx^2$, where a and b are constants. At the left end, the resistivity is 2.25 $\times$ 10$^{-8} \Omega$ $\cdot$ m, while at the right end it is 8.50 $\times$ 10$^{-8}\Omega$ $\cdot $ m. (a) What is the resistance of this rod? (b) What is the electric field at its midpoint if it carries a 1.75-A current? (c) If we cut the rod into two 75.0-cm halves, what is the resistance of each half?

Transcript

00:02 So in this problem, we have a rod, but the material is very complicated.

00:08 But we know that the resistivity has relation to row equal a plus b times x squared, where x is the distance from the, from some point in the road to one side, to the left side, let's see.

00:24 Okay, so this is x.

00:28 And we also know that on the left side, the resistivity is, is the resistivity r l, sorry, roll l, equal 2 .25 times 10 to the negative 8 meter.

00:45 And on the right side, roll r equal 8 .5 times 10 to the negative 8 omel meter.

00:57 So by using this to end, by using these two boundary conditions, we can construct two equations.

01:04 So the first one is a plus b times 0 equal 2 .25 times 10 to the negative 8th.

01:15 And second one is a a plus b times length, which is 1 .5 squared, equal 8 .5 times 10 to the negative 8.

01:27 So it can obtain the solution of a and a b.

01:32 So a equal 2 .25 times 10 to the negative 8th omometer.

01:40 And b is a 2 .78 times 10 to negative home per meter.

01:48 Okay.

01:51 So now that we have the parameter a and the b, we can find out the resistance of the whole rod.

02:00 So if we just look at a small pieces of the rod, we can see that the dr means the resistance of this small pieces equal row times dx over a, right? and the role has expression a plus bx squared and times dx over a right so the total resistance equal integral d r always say it equal uh in the growth of d x so x ranges from zero to l a time a plus bx squared and d x over a okay so uh because we already have the value of a and the b and a b and a has expression pi times r squared right so radius is given the problem so you can find out the the value of this integral and i found it is 1 .71 times 10 to negative 4 forms okay and in part b i want to find out the the electric field at the middle point so basically we just look at this small piece at the middle point okay then for this small piece we have the potential difference dv equal i times d r right so current is known and d r is the resistance of this small piece so d r has expression right here so this gives us i times row times d x over a and a row has expression i times a plus bx squared dx over a right so the electro field the magnitude of the electro field is defined to be dv over d x in this case, right? so if you look at this expression again, so this is dv, and on the right hand side, there contains dx, right? so you see that dv over dx equal i times a plus bx squared and times 1 over a, right? so simply need to plug in the values, and here x equal l over 2, okay? because we're just looking at the middle point...

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