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You apply a potential difference of \(4.50 \mathrm{~V}\) between the ends of a wire that is \(2.50 \mathrm{~m}\) in length and \(0.654 \mathrm{~mm}\) in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?

Short Answer

Expert verified
The resistivity of the wire is \(2.04 * 10^{-8} Ohm.m\)

Step by step solution

01

Calculate Resistance Using Ohm's Law

Ohm's law states that the current through a conductor between two points is directly proportional to the voltage across the two points. The formula for Ohm's Law is \(I = V/R\), where I is the current, V is the potential difference or voltage, and R is the resistance. We will rearrange this formula to solve for R: \(R = V/I\). Then we plug in the given values: \(R = 4.5V / 17.6A = 0.255682 Ohms\).
02

Convert radius to meters

The radius of the wire is given in millimeters but the standard SI unit for measuring length in such physical situations is meter. So, the radius needs to be converted into meters. This can be done by multiplying the given value by \(10^-3\). So, \(0.654mm = 0.654 * 10^{-3} m = 0.000654 m\).
03

Calculate Resistivity

Resistivity can be calculated from resistance, length and radius of the wire using the formula \(Resistivity = Resistance * (Area / Length)\), where Area can be replaced by \( \pi r^2\), as it is the cross-sectional area of the wire, and r is its radius. Substituting known values in the formula: \(Resistivity = 0.255682 Ohms * (\pi * (0.000654m)^2 / 2.5m)\), and with calculation, the resistivity is found to be \(2.04 * 10^{-8} Ohm.m\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ohm's Law
Understanding Ohm's Law is crucial for anyone studying electricity. It's the foundation upon which the functionality of many electrical circuits is based. Simply put, this law states that the current flowing through a conductor between two points is directly proportional to the voltage across those two points and inversely proportional to the resistance of the conductor. The mathematical representation is given by the formula:

\textbf{Ohm's Law:} \( I = \frac{V}{R} \)
where \( I \) is the electric current in amperes (A), \( V \) is the potential difference in volts (V), and \( R \) is the resistance in ohms (Ω).
To put this into context with our exercise, knowing the potential difference and the current allowed us to rearrange the equation to solve for resistance. This law is not just a formula; it's a window into understanding how electrical components interact in a circuit. Thinking of it visually can help; imagine voltage as the pressure pushing water through a pipe, resistance as the narrowness of the pipe, and current as the flow of water.
Electrical Resistance
Electrical resistance is a fundamental concept, which tells us how much a material opposes the flow of electric current. The higher the resistance, the harder it is for current to flow. Resistance can be likened to friction in mechanical systems—it's what hampers the smooth flow of current.

In our example, we calculated the resistance of a wire. The formula for calculating resistance when the resistivity, length, and cross-sectional area are known is:
\textbf{Resistance formula:} \( R = \rho \frac{l}{A} \)
where \( \rho \) is resistivity, \( l \) is the length of the conductor, and \( A \) is the cross-sectional area of the conductor. With our wire, the length and the calculated resistance were given, so we could use them in a rearranged version of the resistivity formula to find the resistivity. It's important to visualize that resistance in a wire is like a hurdle to the current: the longer the wire, or the thinner it is, the higher the resistance—and thereby, the lower the current for a given voltage.
Conductivity and Resistivity
Conductivity and resistivity are two sides of the same coin. While conductivity measures how easily electric charges can pass through a material, resistivity is all about the material's opposition to those charges. High resistivity means low conductivity and vice versa. Conductivity is represented by the symbol \( \sigma \), and resistivity by \( \rho \).

\textbf{Resistivity formula:} \( \rho = R\frac{A}{l} \)
In our wire problem, we used the known resistance and the physical dimensions of the wire to find its resistivity. The formula takes into account the entire physical structure of the wire—which is why we needed the wire's length and the area of its cross-section (calculated from its radius). When students struggle with this concept, envisioning the material as a busy road can be helpful; resistivity is akin to the number of lanes—if there are more lanes (higher area), traffic (current) flows better, leading to lower resistivity. Every material has a characteristic resistivity, which is a measure of its ability to conduct electricity, essential in designing electrical circuits and components.

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Most popular questions from this chapter

\(A 3.00 \mathrm{~m}\) length of copper wire at \(20^{\circ} \mathrm{C}\) has a \(1.20-\mathrm{m}\) -long section with diameter \(1.60 \mathrm{~mm}\) and a \(1.80-\mathrm{m}\) -long section with diameter \(0.80 \mathrm{~mm}\). There is a current of \(2.5 \mathrm{~mA}\) in the 1.60 -mm-diameter section. (a) What is the current in the 0.80 -mm-diameter section? (b) What is the magnitude of \(\vec{E}\) in the 1.60 -mm-diameter section? (c) What is the magnitude of \(\vec{E}\) in the 0.80 -mm-diameter section? (d) What is the potential difference between the ends of the \(3.00 \mathrm{~m}\) length of wire?

The current in a wire varies with time according to the relationship \(I=55 \mathrm{~A}-\left(0.65 \mathrm{~A} / \mathrm{s}^{2}\right) t^{2} .\) (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t=0\) and \(t=8.0 \mathrm{~s} ?(\mathrm{~b}) \mathrm{What}\) constant current would transport the same charge in the same time interval?

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

In Fig. 29.23 the capacitor plates have area \(5.00 \mathrm{~cm}^{2}\) and separation \(2.00 \mathrm{~mm} .\) The plates are in vacuum. The charging current \(i_{\mathrm{C}}\) has a constant value of \(1.80 \mathrm{~mA}\). At \(t=0\) the charge on the plates is zero. (a) Calculate the charge on the plates, the electric field between the plates, and the potential difference between the plates when \(t=0.500 \mu \mathrm{s}\). (b) Calculate \(d E / d t,\) the time rate of change of the electric field between the plates. Does \(d E / d t\) vary in time? (c) Calculate the displacement current density \(j_{\mathrm{D}}\) between the plates, and from this the total displacement current \(i_{\mathrm{D}}\). How do \(i_{\mathrm{C}}\) and \(i_{\mathrm{D}}\) compare?

A material with resistivity \(\rho\) is formed into a cylinder of length \(L\) and outer radius \(r_{\text {outer }}\). A cylindrical core with radius \(r_{\text {inner }}\) is removed from the axis of this cylinder and filled with a conducting material, which is attached to a wire. The outer surface of the cylinder is coated with a conducting material and attached to another wire. (a) If the second wire has potential \(V\) greater than the first wire, in what direction does the local electric field point inside of the cylinder? (b) The magnitude of this electric field is \(c / r,\) where \(c\) is a constant and \(r\) is the distance from the axis of the cylinder. Use the relationship \(V=\int \overrightarrow{\boldsymbol{E}} \cdot d \overrightarrow{\boldsymbol{l}}\) to determine the constant \(c .(\mathrm{c})\) What is the resistance of this device? (d) A \(1.00-\mathrm{cm}\) -long hollow cylindrical resistor has an inner radius of \(1.50 \mathrm{~mm}\) and an outer radius of \(3.00 \mathrm{~mm} .\) The material is a blend of powdered carbon and ceramic whose resistivity \(\rho\) may be altered by changing the amount of carbon. If this device should have a resistance of \(6.80 \mathrm{k} \Omega,\) what value of \(\rho\) should be selected?

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