A 1.50 m cylinder of radius 1.10 cm is made of a complicated mixture of materials. Its resisti

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

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

Key Concepts

Series Combination of Resistive Segments

Dividing a resistor into segments results in a series configuration, where the total resistance is the sum of the individual resistances of each segment. This concept is essential when analyzing composite or partitioned systems, as it allows the resistance of each segment to be calculated separately and then combined to understand the overall behavior of the circuit.

Variable Resistivity

Resistivity is a material property that quantifies how strongly a given material opposes the flow of electric current. In problems where resistivity is not constant but depends on position, it becomes necessary to incorporate its spatial variation directly into the calculations. This allows us to model materials whose composition or structure changes along their length, influencing the overall resistance and the distribution of electric fields within them.

Integration of Differential Resistance

When dealing with an object whose resistivity varies with position, the overall resistance is determined by integrating the infinitesimal resistance contributions along its length. The differential resistance is given by dR = ?(x)dx/A, where A is the constant cross-sectional area. By setting up and evaluating this integral over the length of the material, one can compute the total resistance, a method that is central in applying calculus techniques to problems in electromagnetism.

Cross-Sectional Area

The cross-sectional area of a conductor plays a crucial role in determining its resistance, since resistance is inversely proportional to this area. For objects with simple geometries, like cylinders, the area is calculated using the standard geometric formula. This concept is key in relating the physical dimensions of a conductor to its electrical properties.

Electric Field in a Conductor

The electric field within a conductor is related to the local current density and the material's resistivity by the relation E = ?J. In cases where the resistivity varies spatially, the electric field may also vary, and specific points within the conductor (such as the midpoint) need to be analyzed using the local values of resistivity alongside the imposed current.

Transcript

00:01 Okay, so we know the resistance can be equal to integral range from 0 to l times row x over a dx.

00:08 L here is the length of the cylinder.

00:14 And the row x is the function of the resistivity, respect with the x, which is equal to a plus bx square.

00:22 So if we do some arrangement here, eventually you have r is equal to 1 over a times the integral.

00:33 So we know the anti -derivative for a plus bx square is a x plus bxq over 3.

00:41 So now we can plug in the values for the integral.

00:45 Eventually, we'll have the resistance is equal to l over a times lowercase a plus bl square over 3.

00:54 So we know the area is equal to pi r square, which is equal to pi times 1 .1 times 10 to the power of negative meter to the power 2.

01:01 Since the radius is given as 1 .1 centimeter and 1 .1 centimeter is equal to 1 .1 times 10 to the power of negative 2 meter.

01:09 And this will give us the area is equal to 1 .21 pi tens tend to the power of a negative 4 meter square.

01:15 And we know the length of the cylinder is given as 1 .5 meter.

01:20 And a is the value for the resistivity at the left end, which is equal to 2 .25 times 10 to the power of negative 8 omega times meter.

01:30 For the resistivity at the right end, we need to use the function here.

01:38 So the resistivity at the right end is actually equal to row l in this case.

01:48 So if we do some arrangement here, we have b is equal to row l minus a over l square.

01:53 And now if we plug in the values, we have b is equal to 2 .7 a times 10 to the power of negative a omega times meter.

02:01 So now let's plug in the values to determine the resistance.

02:04 Which is equal to 1 .50 meter over 1 .21 pi times 10 to the power of negative 4 meter square and then multiplied by 2 .25 times 10 to the power of negative 8 omega meter plus 2 .78 times 10 to the power of negative 8 omega meter plus 2 .78 times 10 to the power of negative 8 omega meter plus plus 2 .78 times 10 to the power of of negative 8 omega meter multiplied by 1 .5 meter square over 3...

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