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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 ductile metal wire has resistance \(R\). What will be the resistance of this wire in terms of \(R\) if it is stretched to three times its original length, assuming that the density and resistivity of the material do not change when the wire is stretched? (Hint: The amount of metal does not change, so stretching out the wire will affect its cross-sectional area.)

The resistivity of a semiconductor can be modified by adding different amounts of impurities. A rod of semiconducting material of length \(L\) and cross- sectional area \(A\) lies along the \(x\) -axis between \(x=0\) and \(x=L\). The material obeys Ohm's law, and its resistivity varies along the rod according to \(\rho(x)=\rho_{0} \exp (-x / L) .\) The end of the rod at \(x=0\) is at a potential \(V_{0}\) greater than the end at \(x=L\). (a) Find the total resistance of the rod and the current in the rod. (b) Find the electric-field magnitude \(E(x)\) in the rod as a function of \(x\). (c) Find the electric potential \(V(x)\) in the rod as a function of \(x\). (d) Graph the functions \(\rho(x), E(x),\) and \(V(x)\) for values of \(x\) between \(x=0\) and \(x=L\)

A battery-powered global positioning system (GPS) receiver operating on \(9.0 \mathrm{~V}\) draws a current of 0.13 A. How much electrical energy does it consume during 30 minutes?

A coil \(4.00 \mathrm{~cm}\) in radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to \(B=(0.0120 \mathrm{~T} / \mathrm{s}) t+\left(3.00 \times 10^{-5} \mathrm{~T} / \mathrm{s}^{4}\right) t^{4} .\) The coil is connected to a \(600 \Omega\) resistor, and its plane is perpendicular to the magnetic field. You can ignore the resistance of the coil. (a) Find the magnitude of the induced emf in the coil as a function of time. (b) What is the current in the resistor at time \(t=5.00 \mathrm{~s} ?\)

A metal ring \(4.50 \mathrm{~cm}\) in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of \(1.12 \mathrm{~T}\) between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at \(0.250 \mathrm{~T} / \mathrm{s}\). (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

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