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The potential difference between points in a wire 75.0 \(\mathrm{cm}\) apart is 0.938 \(\mathrm{V}\) when the current density is \(4.40 \times 10^{7} \mathrm{A} / \mathrm{m}^{2}\) . What are (a) the magnitude of \(\overrightarrow{\boldsymbol{E}}\) in the wire and \((\mathrm{b})\) the resistivity of the material of which the wire is made?

Short Answer

Expert verified
(a) \( \overrightarrow{E} \approx 1.25 \text{ V/m} \), (b) resistivity \( \approx 2.84 \times 10^{-8} \text{ Ωm}. \)

Step by step solution

01

Understanding the Problem

We are given a potential difference (voltage) across a length of wire, a current density, and the length of the wire. From this, we need to find the electric field magnitude \( \overrightarrow{E} \) and the resistivity of the material.
02

Formula for Electric Field

The electric field \( \overrightarrow{E} \) in a material is related to the potential difference (voltage) \( V \) and the distance \( L \) over which this voltage is applied: \[ \overrightarrow{E} = \frac{V}{L}. \]Here, \( V = 0.938 \text{ V} \) and \( L = 75.0 \text{ cm} = 0.75 \text{ m}.\)
03

Calculate \( \overrightarrow{E} \)

Substitute the values into the formula for electric field: \[ \overrightarrow{E} = \frac{0.938}{0.75} \approx 1.25 \text{ V/m}. \]
04

Understanding Resistivity

Resistivity \( \rho \) can be calculated using the relation between resistivity, electric field \( \overrightarrow{E} \), and current density \( J \):\[ \overrightarrow{E} = \rho J. \] Thus, \( \rho = \frac{\overrightarrow{E}}{J}. \)
05

Calculate Resistivity

Given that \( \overrightarrow{E} = 1.25 \text{ V/m} \) and \( J = 4.40 \times 10^{7} \text{ A/m}^2 \), use the formula for resistivity:\[ \rho = \frac{1.25}{4.40 \times 10^{7}} \approx 2.84 \times 10^{-8} \text{ Ωm}. \]

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Key Concepts

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

Current Density
Current density is a crucial concept in understanding how electric current flows through a material. It helps us understand the distribution of electric current in a given area. In simple terms, current density is the amount of electric current flowing per unit area of a cross-section.

To visualize it, consider the current as the total number of electrical charges passing through a wire. Current density tells us how closely packed these charges are within the wire. If the current is the same, but the wire's thickness varies, the current density will change too. It is akin to how traffic density works — on a narrow road, more cars (charge carriers) will make traffic denser.

The formula for current density, denoted as \(J\), is given by:
  • \(J = \frac{I}{A}\)
where \(I\) is the current in amperes, and \(A\) is the cross-sectional area in square meters. Architects of electrical systems often consider current density to ensure systems can handle the electrical load without overheating.
Resistivity
Resistivity is an intrinsic property of a material that quantifies how strongly that material opposes the flow of electric current. It is an essential concept when determining how a material will conduct electricity.

Let’s imagine resistivity as the frictional force acting against the flow of electricity. Materials with high resistivity, such as rubber, impede the flow significantly, making them good insulators. On the contrary, materials like copper have low resistivity, allowing electrons to move freely, making them excellent conductors.

Resistivity, symbolized by \(\rho\), can be calculated using the formula:
  • \(\rho = \frac{\overrightarrow{E}}{J}\)
where \(\overrightarrow{E}\) is the electric field and \(J\) is the current density. In practical applications, understanding a material's resistivity helps in making crucial decisions about which materials to use for specific parts of electrical circuits to optimize performance and safety.
Potential Difference
Potential difference, often referred to as voltage, is what drives current through a circuit. Think of it as the force that pushes electric charges through a conductor.

A simple analogy can be found in water flow: if we think of electric current as the flow of water, the potential difference is like the water pressure that pushes the water through pipes. Without this force, there's no movement of charges, just as without pressure, water wouldn't flow.

The potential difference \(V\) between two points in a circuit is defined by the work done \(W\) in moving a unit charge \(Q\) between the two points, expressed as:
  • \(V = \frac{W}{Q}\)
This tells us the energy needed to move each unit of charge between two points. A higher potential difference means a stronger "push," resulting in more current through the circuit. Understanding potential difference helps in designing circuits with the desired performance and selecting appropriate power supplies.

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Most popular questions from this chapter

A carbon resistor is to be used as a thermometer. On a winter day when the temperature is \(4.0^{\circ} \mathrm{C}\) , the resistance of the carbon resistor is 217.3\(\Omega\) . What is the temperature on a spring day when the resistance is 215.8\(\Omega ?\) (Take the reference temperature \(T_{0}\) to be \(4.0^{\circ} \mathrm{C} . )\)

A cylindrical copper cable 1.50 \(\mathrm{km}\) long is connected across a 220.0 - V potential difference. (a) What should be its diameter so that it produces heat at a rate of 50.0 \(\mathrm{W}\) ? (b) What is the electric field inside the cable under these conditions?

The current-voltage relationship of a semiconductor diode is given by $$ I=I_{\mathrm{S}}\left[\exp \left(\frac{e V}{k T}\right)-1\right] $$ where \(I\) and \(V\) are the current through and the voltage across the diode, respectively. Is a constant characteristic of the device, \(e\) is the magnitude of the electron charge, \(k\) is the Boltzmann constant, and \(T\) is the Kelvin temperature. Such a diode is connected in series with a resistor with \(R=1.00 \Omega\) and a battery with \(\mathcal{E}=2.00 \mathrm{V} .\) The polarity of the battery is such that the current through the diode is in the forward direction (Fig. \(25.45 ) .\) The battery has negligible internal resistance. (a) Obtain an equation for \(V .\) Note that you cannot solve for \(V\) algebraically. (b) The value of \(V\) must be obtained by using a numerical method. One approach is to try a value of \(V\) , see how the left-and right-hand sides of the equation compare for this \(V\) , and use this to refine your guess for \(V\) . Using \(I_{S}=1.50 \mathrm{mA}\) and \(T=293 \mathrm{K},\) obtain a solution (accurate to three significant figures) for the voltage drop \(V\) across the diode and the current \(I\) through it.

On your first day at work as an electrical technician, you are asked to determine the resistance per meter of a long piece of wire. The company you work for is poorly equipped. You find a battery, a voltmeter, and an ammeter, but no meter for directly measuring resistance (an ohmmeter). You put the leads from the voltmeter across the terminals of the battery, and the meter reads 12.6 \(\mathrm{V}\) . You cut off a 20.0 -m length of wire and connect it to the battery, with an ammeter in series with it to measure the current in the wire. The ammeter reads 7.00 A. You then cut off a 40.0 -m length of wire and connect it to the battery, again with the ammeter in series to measure the current. The ammeter reads 4.20 \(\mathrm{A}\) . Even though the equipment you have available to you is limited, your boss assures you of its high quality: The ammeter has very small resistance, and the voltmeter has very large resistance. What is the resistance of 1 meter of wire?

A \(12.6-\mathrm{V}\) car battery with negligible internal resistance is connected to a series combination of a \(3.2-\Omega\) resistor that obeys Ohm's law and a thermistor that does not obey Ohm's law but instead has a current- voltage relationship \(V=\alpha I+\beta I^{2},\) with \(\alpha=3.8 \Omega\) and \(\beta=1.3 \Omega / \mathrm{A}\) . What is the current through the \(3.2-\Omega\) resistor?

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