/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A steady current of \(2 \mathrm{... [FREE SOLUTION] | 91影视

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Chapter 34: Problem 1

A steady current of \(2 \mathrm{~A}\) in a coil of 400 turns causes a flux of \(10^{-4}\) \(\mathrm{Wb}\) to link (pass through) the loops of the coil. Compute ( \(a\) ) the average back emf induced in the coil if the current is stopped in \(0.08 \mathrm{~s},(b)\) the inductance of the coil, and \((c)\) the energy stored in the coil. (a) \(|\varepsilon|=N\left|\frac{\Delta \Phi_{M}}{\Delta t}\right|=400 \frac{\left(10^{-4}-0\right) \mathrm{Wb}}{0.08 \mathrm{~s}}=0.5 \mathrm{~V}\) (b) \(|\boldsymbol{E}|=N\left|\frac{\Delta i}{\Delta t}\right| \quad\) or \(\quad L=\left|\frac{\delta \Delta t}{\Delta i}\right|=\frac{(0.5 \mathrm{~V})(0.08 \mathrm{~s})}{(2-0) \mathrm{A}}=0.02 \mathrm{H}\) (c) Energy \(=\frac{1}{2} L I^{2}=\frac{1}{2}(0.02 \mathrm{H})(2 \mathrm{~A})^{2}=0.04 \mathrm{~J}\)

Short Answer

Expert verified
(a) 0.5 V, (b) 0.02 H, (c) 0.04 J

Step by step solution

01

Calculate the Average Back EMF Induced

The formula for average back EMF induced in a coil is given by \[ |\varepsilon| = N \left| \frac{\Delta \Phi_{M}}{\Delta t} \right| \]where \( N \) is the number of turns, \( \Delta \Phi_M \) is the change in magnetic flux, and \( \Delta t \) is the change in time. Substitute the given values:\[ |\varepsilon| = 400 \frac{(10^{-4}-0) \mathrm{Wb}}{0.08 \mathrm{~s}} = 0.5 \mathrm{~V} \]
02

Calculate the Inductance of the Coil

Inductance \( L \) of the coil can be calculated using the change in EMF over the change in current:\[ L = \left| \frac{\Delta t \times \varepsilon}{\Delta i} \right| \]Substitute \( \varepsilon = 0.5 \mathrm{~V} \), \( \Delta t = 0.08 \mathrm{~s} \) and \( \Delta i = 2 \mathrm{~A} \):\[ L = \frac{(0.5 \mathrm{~V})(0.08 \mathrm{~s})}{2 \mathrm{~A}} = 0.02 \mathrm{~H} \]
03

Calculate the Energy Stored in the Coil

The energy stored in the coil is given by the formula:\[ \text{Energy} = \frac{1}{2} L I^2 \]Plugging in the values \( L = 0.02 \mathrm{~H} \) and \( I = 2 \mathrm{~A} \):\[ \text{Energy} = \frac{1}{2} (0.02 \mathrm{~H}) (2 \mathrm{~A})^2 = 0.04 \mathrm{~J} \]

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Key Concepts

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

Back EMF Calculation
When a current passing through a coil is suddenly interrupted, an induced electromotive force, known as back EMF, is generated. This is a result of the rapid change in magnetic flux within the coil's loops. The back EMF is calculated using the formula: \[|\varepsilon| = N \left| \frac{\Delta \Phi_{M}}{\Delta t} \right|\] Where:
  • \(N\) = Number of turns in the coil
  • \(\Delta \Phi_{M}\) = Change in magnetic flux linkage
  • \(\Delta t\) = Time period over which the change happens
In this specific exercise, with \(400\) turns and a flux change from \(10^{-4} \mathrm{Wb}\), the back EMF is calculated as \(0.5 \mathrm{V}\). Understanding the back EMF is crucial as it acts against the applied voltage and affects the performance of electrical motors and transformers.
Magnetic Flux Linkage
Magnetic flux linkage is the measure of the total magnetic field which is saying 'hello' to the cross-sectional area of a coil. It's the product of the magnetic flux and the number of turns in the coil, expressed as \(N \times \Phi_{M}\). Let's break it down:
  • \(\Phi_{M}\) stands for magnetic flux, representing the strength and number of magnetic field lines passing through an area
  • Linkage considers all loops or turns in the coil, showing combined interaction
In the given problem, the magnetic flux linkage means all 400 coil turns contributing together to the change in flux. This linkage is important for understanding how inductive components work in electrical circuits.
Energy Stored in Inductors
Inductors store energy in the form of a magnetic field created by the flow of electrical current through the coil. The energy stored in an inductor is given by: \[\text{Energy} = \frac{1}{2} L I^2\] Where:
  • \(L\) = Inductance of the coil
  • \(I\) = Current flowing through the coil
In this exercise, the calculated energy is \(0.04 \mathrm{~J}\). This concept of storing energy is crucial in numerous applications like energy transfer in transformers, and in smoothing circuits in power supplies to maintain a constant current flow.
Coil Inductance
Coil inductance, denoted as \(L\), is influential in determining how effectively a coil can store and release energy. It's a measure of a coil's ability to induce an EMF given a change in current. Using the relation: \[L = \left| \frac{\Delta t \times \varepsilon}{\Delta i} \right| \] Where:
  • \(\Delta t\) = Time over which the change occurs
  • \(\varepsilon\) = Induced EMF
  • \(\Delta i\) = Change in current
The inductance in this scenario is \(0.02 \mathrm{~H}\). Coil inductance affects how smoothly electrical currents are managed, with higher values indicating increased ability to store magnetic energy.
Current in Coils
The current flowing through a coil is essential for understanding how inductors work. This current creates a magnetic field within the coil, linking back to magnetic flux and energy storage concepts. The formula helping us visualize the energy stored relates to this current: \[\text{Energy} = \frac{1}{2} L I^2\] In our problem, with a steady current \(I\) of \(2 \mathrm{A}\), it enables the flux linkage and contributes directly to the energy accumulated within the magnetic field. Changes in this current affect the inductor鈥檚 behavior, as seen from stopping the current in this exercise, resulting in the calculated back EMF.

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Most popular questions from this chapter

[II] When a long iron-core solenoid is connected across a 6-V battery, the current rises to \(0.63\) of its maximum value after a time of \(0.75 \mathrm{~s}\). The experiment is then repeated with the iron core removed. Now the time required to reach \(0.63\) of the maximum is \(0.0025 \mathrm{~s}\). Calculate ( \(a\) ) the relative permeability of the iron and \((b) L\) for the aircore solenoid if the maximum current is \(0.5 \mathrm{~A}\).

A coil of resistance \(15 \Omega\) and inductance \(0.60 \mathrm{H}\) is connected to a steady 120 -V power source. At what rate will the current in the coil rise \((a)\) at the instant the coil is connected to the power source, and \((b)\) at the instant the current reaches 80 percent of its maximum value? The effective driving voltage in the circuit is the \(120 \mathrm{~V}\) power supply minus the induced back emf, \(\mathrm{L}(\Delta \mathrm{i} / \Delta t)\). This equals the p.d. in the resistance of the coil: $$ 120 \mathrm{~V}-L \frac{\Delta i}{\Delta t}=i R $$ [This same equation can be obtained by writing the loop equation for the circuit of \(\underline{\text { Fig. } 34-2(a) \text { . In doing so, remember that the }}\) inductance acts as a back emf of value \(\mathrm{L}(\Delta i / \Delta t) .]\) (a) At the first instant, \(i\) is essentially zero. Then $$ \frac{\Delta i}{\Delta t}=\frac{120 \mathrm{~V}}{L}=\frac{120 \mathrm{~V}}{0.60 \mathrm{H}}=0.20 \mathrm{~mA} / \mathrm{s} $$ (b) The current reaches a maximum value of \((120 \mathrm{~V}) / R\) when the current finally stops changing (i.e., when \(\Delta i / \Delta t=0\) ). We are interested in the case when $$ i=(0.80)\left(\frac{120 \mathrm{~V}}{R}\right) $$ Substitution of this value for \(i\) in the loop equation gives $$ 120 \mathrm{~V}-L \frac{\Delta i}{\Delta t}=(0.80)\left(\frac{120 \mathrm{~V}}{R}\right) R $$ from which $$ \frac{\Delta i}{\Delta t}=\frac{(0.20)(120 \mathrm{~V})}{L}=\frac{(0.20)(120 \mathrm{~V})}{0.60 \mathrm{H}}=40 \mathrm{~A} / \mathrm{s} $$

A \(2.0-\mu \mathrm{F}\) capacitor is charged through a \(30-\mathrm{M} \Omega\) resistor by a 45 \(\mathrm{V}\) battery. Find \((a)\) the charge on the capacitor and \((b)\) the current through the resistor, both determined \(83 \mathrm{~s}\) after the charging process starts. The time constant of the circuit is \(R C=60 \mathrm{~s}\). Also, $$ q_{\infty}=V_{\infty} C=(45 \mathrm{~V})\left(2.0 \times 10^{-6} \mathrm{~F}\right)=9.0 \times 10^{-6} \mathrm{C} $$ (a) \(q=q_{\infty}\left(1-e^{-t / R C}\right)=\left(9.0 \times 10^{-5} \mathrm{C}\right)\left(1-e^{-83 / 60}\right)\) But \(\quad e^{-83 / 60}=e^{-1.383}=0.25\) Then substitution gives $$ q=\left(9.0 \times 10^{-5} \mathrm{C}\right)(1-0.25)=67 \mu \mathrm{C} $$ (b) \(i=i_{0} e^{-t / R C}=\left(\frac{45 \mathrm{~V}}{30 \times 10^{6} \Omega}\right)\left(e^{-1.383}\right)=0.38 \mu \mathrm{A}\)

A coil of inductance \(0.20 \mathrm{H}\) and \(1.0-\Omega\) resistance is connected to a constant \(90-\mathrm{V}\) source. At what rate will the current in the coil grow \((a)\) at the instant the coil is connected to the source, and \((b)\) at the instant the current reaches two-thirds of its maximum value?

How much energy is stored in a 0.500-H inductor carrying a current of \(4.80 \mathrm{~A}\) ?
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