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Chapter 25: Problem 7

The current in a wire varies with time according to the relationship \(I = 55 A\) - \(10.65 A/s{^2}2t{^2}\). (a) How many coulombs of charge pass a cross section of the wire in the time interval between \(t =\) 0 and \(t =\) 8.0 s? (b) What constant current would transport the same charge in the same time interval?

Short Answer

Expert verified
212.8 C pass the cross-section, and the equivalent constant current is 26.6 A.

Step by step solution

01

Understand the Problem

We are given a current function that varies with time: \( I(t) = 55 - 10.65t^2 \). We need to find the total charge that passes through a cross-section of the wire between \( t = 0 \) and \( t = 8.0 \) seconds, and then determine the constant current that would carry the same charge over this time period.
02

Set Up the Integral for Charge

Charge (\( Q \)) passing a cross-section of a wire is given by the integral of the current over time: \[ Q = \int_{t_1}^{t_2} I(t) \, dt \]Where \( t_1 = 0 \) s and \( t_2 = 8.0 \) s. Substituting the expression for \( I(t) \), we have:\[ Q = \int_{0}^{8} (55 - 10.65t^2) \, dt \]
03

Perform the Integration

Integrate the function:\[ Q = \int (55 - 10.65t^2) \, dt \]This can be split into the difference of two integrals:\[ Q = \int 55 \, dt - \int 10.65t^2 \, dt \]Calculating each:\[ \int 55 \, dt = 55t \]and\[ \int 10.65t^2 \, dt = \frac{10.65}{3}t^3 \]Thus, the integrated function is:\[ Q = 55t - \frac{10.65}{3}t^3 \].
04

Evaluate the Definite Integral

Evaluate the integrated function from \(0\) to \(8\) seconds:\[ Q = \left[55t - \frac{10.65}{3}t^3\right]_0^8 \]This becomes:\[ Q = \left(55(8) - \frac{10.65}{3}(8)^3\right) - \left(55(0) - \frac{10.65}{3}(0)^3\right) \]Compute the values:\[ Q = \left(440 - 227.2\right) - 0 \]Thus, \( Q = 212.8 \) coulombs.
05

Calculate the Constant Current

To find the constant current \( I_c \) that transports the same charge in the same time (8 seconds), use the formula:\[ I_c = \frac{Q}{t} \]Substitute \( Q = 212.8 \) C and \( t = 8 \) s:\[ I_c = \frac{212.8}{8} = 26.6 \, A \].

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Key Concepts

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

Time-Varying Current
A time-varying current is one where the amount of electrical current changes as time progresses. In the exercise, we have a current equation: \( I(t) = 55 - 10.65t^2 \). This means that initially, when \( t = 0 \), the current is 55 Amps, but as time increases, the current decreases due to the \(-10.65t^2\) term. The "55" represents the initial current, and the "\(-10.65t^2\)" describes how that current decreases with time, showing that it decreases quadratically. Understanding time-varying currents is crucial in environments where current is not constant, which often reflects real-world applications such as in electronics and circuit design.
Integration in Physics
Integration in physics is a fundamental tool used to calculate quantities like displacement, charge, or energy, over a specified interval. When working with currents, integration helps us find the total charge that flows over a certain period. In the problem, we need to integrate the current function over the time interval from 0 to 8 seconds to determine the total charge transfer. The process involves breaking down complex equations into simpler parts, which can be easily integrated. This technique is essential in physics to transition from rate-based formulas, like current, to total quantities, such as charge.
Definite Integral
A definite integral is used to calculate the exact area under a curve from one point to another, which, in context, helps determine things like total charge. For the exercise, the definite integral of the current function \( I(t) = 55 - 10.65t^2 \) is calculated from 0 to 8 seconds. The result gives the total charge passing through a certain section of the wire. The definite integral includes limits at both ends, \( t_1 \) and \( t_2 \), signifying the precise timeframe over which the function is evaluated. This is important in physics calculations where precise initial and final conditions are necessary.
Constant Current
Constant current is much simpler compared to varying currents; it does not change over time. It is equivalent to the average value of the varying current over a specific period, transporting the same amount of charge in that timeframe. After determining the total charge in the exercise, the constant current needed to carry the same charge over 8 seconds was calculated. This involved using the formula \( I_c = \frac{Q}{t} \), where \( Q \) is the charge and \( t \) is the time interval. Understanding this comparison helps comprehend how varying and constant currents can achieve the same effect under different circumstances.

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Most popular questions from this chapter

An external resistor with resistance \(R\) is connected to a battery that has emf \(\varepsilon\) and internal resistance \(r\). Let \(P\) be the electrical power output of the source. By conservation of energy, \(P\) is equal to the power consumed by \(R\). What is the value of \(P\) in the limit that \(R\) is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when \(R = r\). What is this maximum \(P\) in terms of \(\varepsilon\) and \(r\)? (d) A battery has \(\varepsilon\) = 64.0 V and \(r =\) 4.00 \(\Omega\). What is the power output of this battery when it is connected to a resistor \(R\), for \(R =\) 2.00 \(\Omega\), \(R =\) 4.00 \(\Omega\), and \(R =\) 6.00 \(\Omega\) ? Are your results consistent with the general result that you derived in part (b)?

A 25.0-\(\Omega\) bulb is connected across the terminals of a 12.0-V battery having 3.50\(\Omega\) of internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

A copper wire has a square cross section 2.3 mm on a side. The wire is 4.0 m long and carries a current of 3.6 A. The density of free electrons is 8.5 \(\times\) 10\(^{28}\)/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?

Copper has \(8.5 \times 10^{28}\) free electrons per cubic meter. A 71.0-cm length of 12-gauge copper wire that is 2.05 mm in diameter carries 4.85 A of current. (a) How much time does it take for an electron to travel the length of the wire? (b) Repeat part (a) for 6-gauge copper wire (diameter 4.12 mm) of the same length that carries the same current. (c) Generally speaking, how does changing the diameter of a wire that carries a given amount of current affect the drift velocity of the electrons in the wire?

A wire 6.50 m long with diameter of 2.05 mm has a resistance of 0.0290 \(\Omega\). What material is the wire most likely made of?
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