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A more general definition of the temperature coefficient of resistivity is $$\alpha=\frac{1}{\rho} \frac{d \rho}{d T}$$ where \(\rho\) is the resistivity at temperature \(T .\) (a) Assuming that \(\alpha\) is constant, show that $$ \rho=\rho_{0} e^{\alpha\left(T-T_{0}\right)} $$ where \(\rho_{0}\) is the resistivity at temperature \(T_{0} .\) (b) Using the series expansion \(e^{x} \approx 1+x\) for \(x<<1,\) show that the resistivity is given approximately by the expression \(\rho=\rho_{0}\left[1+\alpha\left(T-T_{0}\right)\right]\) for \(\alpha\left(T-T_{0}\right)<<1\)

Short Answer

Expert verified
\(\rho = \rho_0 e^{\alpha(T - T_0)}\) is derived by integrating the differential equation assuming constant \(\alpha\), and \(\rho = \rho_0 (1 + \alpha(T - T_0))\) is obtained using a first-order approximation for small temperature changes.

Step by step solution

01

Express the Differential Equation

Given the temperature coefficient of resistivity, \(\alpha\), as \(\alpha = \frac{1}{\rho} \frac{d\rho}{dT}\). Since \(\alpha\) is constant, we can rewrite the equation as \(d\rho = \alpha \rho dT\) and separate the variables.
02

Integrate Both Sides

Integrate the left side with respect to \(\rho\) and the right side with respect to \(T\) to obtain \(\int \frac{1}{\rho}d\rho = \int \alpha dT\). The result is \(\ln(\rho) = \alpha T + C\), where \(C\) is the constant of integration.
03

Solve for the Integration Constant, C

Exponentiate both sides to remove the natural logarithm, obtaining \(\rho = e^{\alpha T + C}\). Since \(C\) is a constant, represent it as \(e^C\) for simplicity. Let \(\rho_0 = e^C\) be the resistivity at reference temperature \(T_0\), and substitute it in, yielding \(\rho = \rho_0 e^{\alpha T}\).
04

Adjust the Exponent for the Reference Temperature

Add and subtract \(\alpha T_0\) to the exponent in the formula from Step 3 to match the form in the exercise, giving \(\rho = \rho_0 e^{\alpha T - \alpha T_0} = \rho_0 e^{\alpha(T - T_0)}\).
05

Expand the Exponential Function Using the Series Expansion

Use the series expansion \(e^x \approx 1 + x\) which is valid for \(x << 1\) to approximate \(\rho = \rho_0 e^{\alpha(T - T_0)}\).
06

Apply the Series Expansion to the Resistivity Expression

By plugging the condition \(\alpha(T - T_0) << 1\) into our approximation, we obtain \(\rho = \rho_0 (1 + \alpha(T - T_0))\). This expression gives an approximate linear relationship between resistivity and temperature when the change in temperature is small.

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Key Concepts

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

Resistivity and Temperature Relation
Understanding the relationship between resistivity and temperature is crucial in the field of material science and electronics. Resistivity, denoted as \(\rho\), is a material property that indicates how strongly a material opposes the flow of electric current. One interesting aspect of resistivity is that it generally changes with temperature.

As indicated in the exercise, the temperature coefficient of resistivity, \(\alpha\), is defined as \(\alpha = \frac{1}{\rho} \frac{d\rho}{dT}\), representing the rate of change of resistivity with respect to temperature. Assuming \(\alpha\) is constant, the relationship between resistivity \(\rho\) and temperature \(T\) can be described by an exponential function: \(\rho = \rho_{0} e^{\alpha(T-T_{0})}\), where \(\rho_{0}\) is the resistivity at a reference temperature \(T_{0}\).

Practical Implications

For engineers and scientists, this relationship is vital when designing electronic components that must operate reliably across a range of temperatures. Materials with low temperature coefficients are often chosen for stable electrical performance, while those with high coefficients might be used in temperature sensors or for applications where temperature compensation is required.
Differential Equations in Physics
Differential equations are mathematical tools for modeling relationships where a quantity depends on its own rate of change. In physics, differential equations are the backbone of dynamic systems' analysis. The temperature coefficient of resistivity problem involves a first-order differential equation, \(\frac{d\rho}{dT} = \alpha \rho\), which expresses that the rate of change of resistivity with temperature is proportional to the resistivity itself.

Solving this kind of differential equation typically involves separating variables and integrating both sides, as seen in the step-by-step solution. The integration leads to a natural logarithm, \(\ln(\rho) = \alpha T + C\), which upon exponentiation gives the exponential form of the solution.

Applications

Such equations are not unique to resistivity; they appear in myriad applications, from the decay of radioactive materials to the growth of populations. Mastering the differential equations is, therefore, a staple for students aiming to delve into advanced studies in physics and related engineering fields.
Exponential and Logarithmic Functions in Physics
Exponential and logarithmic functions play significant roles in various physical phenomena. In the context of resistivity, an exponential function describes how resistivity changes with temperature, leading to a non-linear relationship. This nonlinear change is often observable over a wide range of temperatures.

The natural logarithm arises when solving the differential equation that models the resistivity's temperature dependency. By exponentiating both sides after integrating, we transition from a logarithmic equation to an exponential one, highlighting one of the most powerful tools in the physicist's arsenal—the ability to switch between different mathematical representations to find solutions.

Approximations in Physics

Frequently, exponential relationships are approximated using a linear approach, as shown by the series expansion \(e^x \approx 1 + x\) for small \(x\). This helps simplify complex exponential relationships in physics when the conditions permit, making it easier to predict and analyze the behavior of systems under small perturbations, such as slight changes in temperature around a set point.

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The quantity of charge \(q\) (in coulombs) that has passed through a surface of area 2.00 \(\mathrm{cm}^{2}\) varies with time according to the equation \(q=4 t^{3}+5 t+6,\) where \(t\) is in seconds. (a) What is the instantaneous current through the surface at \(t=1.00 \mathrm{s}\) ? (b) What is the value of the current density?

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