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Chapter 25: Problem 11

A metal wire has a circular cross section with radius \(0.800 \mathrm{~mm}\) You measure the resistivity of the wire in the following way: You connect one end of the wire to one terminal of a battery that has emf \(12.0 \mathrm{~V}\) and negligible internal resistance. To the other terminal of the battery you connect a point along the wire so that the length of wire between the battery terminals is \(d\). You measure the current in the wire as a function of \(d\). The currents are small, so the temperature change of the wire is very small. You plot your results as \(I\) versus \(1 / d\) and find that the data lie close to a straight line that has slope \(600 \mathrm{~A} \cdot \mathrm{m} .\) What is the resistivity of the material of which the wire is made?

Short Answer

Expert verified
The resistivity of the material of the wire is approximately \(5.04 脳 10^{-8} \Omega . m\).

Step by step solution

01

Understanding the wire parameters

The cross-sectional radius of the wire is given as 0.800 mm. To make computations easier later on, we would convert this to meters. Hence the radius, \( r = 0.800 mm = 0.0008 m \).
02

Use the slope of I versus \(1 / d\) plot

The slope of the I versus \(1 / d\) plot gives us the current per unit length of wire, or \( I / d = 600 A.m \). As it's mentioned in the problem that one end of the wire is connected to a battery of emf 12.0 V, by Ohm's law, we know that \( V = I.R \). From the slope, we have \( I = 600d \). We can substitute \( I \) into Ohm's law to get: \( V = 600d.R \). Dividing both sides by \( 600d \), we obtain \( R = V / (600d) = 0.02 / d \). R is in Ohms.
03

Application of resistivity formula

The resistivity \( \rho \) of a material of a wire can be found using the formula \( R = \rho (l / A) \), where \( l \) is the length of the wire and \( A \) is the cross sectional area of the wire. We know that \( A = \pi r^2 \) and our calculated resistance \( R = 0.02 / d \), substituting these values we get \( R = \rho (d / \pi (0.0008)^2) \). Now, solving for \( \rho \), we get \( \rho = R \pi (0.0008)^2 / d \). Substituting \( R \) from above, we get \( \rho = 0.02 \pi (0.0008)^2 \) which gives the resistivity of the material of the wire in \( \Omega . m \) units.

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Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in the field of electronics and electrical engineering, describing the relationship between voltage (V), current (I), and resistance (R) in an electrical circuit. The law is succinctly expressed by the equation:
\( V = I \times R \).
According to this equation, voltage across an electrical component is directly proportional to the current flowing through it, with the constant of proportionality being the resistance. For instance, if you double the voltage in a circuit while the resistance remains constant, the current will also double.

The practical application of Ohm's Law can be demonstrated through our textbook exercise. When measuring current as a function of different wire lengths (or through varying resistances), we can use Ohm's Law to derive the resistance for each segment of wire. This is crucial for understanding the behavior of electrical circuits and aids in troubleshooting or designing them effectively.
Resistance Calculation
Calculating resistance is a step that often follows understanding Ohm's Law. Resistance in an electrical circuit impedes the flow of electrons, and it can be determined through various physical properties of the material used. The equation used for this purpose is often expressed as:
\( R = \frac{V}{I} \).
In the context of the exercise provided, resistance (R) was calculated using the slope of the current (I) versus the inverse of distance (1/d) plot from the given data. Here, resistance gets derived from the known voltage applied and the particular length of wire being tested by rearranging Ohm's Law to solve for resistance. The technique of plotting I versus 1/d allows for a linear relationship whereby the resistance can be inferred from the slope. This approach is invaluable for materials testing and quality control in manufacturing electronic components.
Conductivity of Materials
The conductivity of a material is essentially the inverse of electrical resistivity. It measures a material's ability to conduct an electric current and is expressed by the Greek letter sigma (\(\sigma\)). A high conductivity means low resistivity, indicating that a material allows electric current to pass through it more easily.

In our textbook exercise, the resistivity (\(\rho\)) of the wire's material was calculated, which is directly related to conductivity (
\(\sigma = 1/\rho\)).
The formula that connects resistivity to the physical dimensions of the wire and its resistance is:
\( R = \rho \frac{l}{A} \),
where 'R' is the resistance, 'l' is the length of the wire, and 'A' is the cross-sectional area. Understanding the material's resistivity and thus its conductivity is crucial in selecting the right materials for electrical wiring and designing efficient electronic systems. It's important to note that resistivity is dependent on the material type and is significantly affected by temperature, which was assumed constant in the given problem for simplicity.

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Most popular questions from this chapter

Displacement Current in a Dielectric. Suppose that the parallel plates in Fig. 29.23 have an area of \(3.00 \mathrm{~cm}^{2}\) and are separated by a 2.50 -mm- thick sheet of dielectric that completely fills the volume between the plates. The dielectric has dielectric constant \(4.70 .\) (You can ignore fringing effects.) At a certain instant, the potential difference between the plates is \(120 \mathrm{~V}\) and the conduction current \(i_{\mathrm{C}}\) equals \(6.00 \mathrm{~mA} .\) At this instant, what are (a) the charge \(q\) on each plate; (b) the rate of change of charge on the plates; (c) the displacement current in the dielectric?

A copper wire has a square cross \(2.3 \mathrm{~mm}\) on a side. The wire is \(4.0 \mathrm{~m}\) long and carries a current of \(3.6 \mathrm{~A}\). The density of free electrons is \(8.5 \times 10^{28} / \mathrm{m}^{3} .\) Find the magnitudes of (a) the current density in the wire and (b) the electric field in the wire. (c) How much time is required for an electron to travel the length of the wire?

Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of \(165.0 \mathrm{~cm}\), but its circumference is decreasing at a constant rate of \(12.0 \mathrm{~cm} / \mathrm{s}\) due to a tangential pull on the wire. The loop is in a constant, uniform magnetic field oriented perpendicular to the plane of the loop and with magnitude \(0.500 \mathrm{~T}\). (a) Find the emf induced in the loop at the instant when \(9.0 \mathrm{~s}\) have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

A single loop of wire with an area of \(0.0900 \mathrm{~m}^{2}\) is in a uniform magnetic field that has an initial value of \(3.80 \mathrm{~T},\) is perpendicular to the plane of the loop, and is decreasing at a constant rate of \(0.190 \mathrm{~T} / \mathrm{s}\)(a) What emf is induced in this loop? (b) If the loop has a resistance of \(0.600 \Omega,\) find the current induced in the loop.

\(\mathrm{A}\) cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity \(Q\) into coulombs. (c) Convert the energy capacity \(U\) into joules. (d) Multiply the charge rating \(Q\) by the potential rating \(V,\) and verify that this is equivalent to the energy capacity \(U\). (e) If the charge \(Q\) were stored on a parallel-plate capacitor with air as the dielectric, at the potential \(V,\) what would be the corresponding capacitance? (f) If the energy in the battery were used to heat \(1 \mathrm{~L}\) of water, estimate the corresponding change in the water temperature? (The heat capacity of water is \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .)\)
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