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Chapter 34: Problem 18

Determine the back emf induced in a coil whose self-inductance is \(8.20 \mathrm{mH}\) when the current through the coil is changing at a constant rate of 100 A per second. [Hint: Use the defining expression for \(L\), Eq. (34.2).]

Short Answer

Expert verified
The back emf induced is \(-0.82 \, \text{V}\).

Step by step solution

01

Understand the Problem

We need to find the back electromotive force (emf) induced in a coil. According to the problem, we have the self-inductance of the coil, \(L = 8.20 \, \text{mH}\), and the rate of change of current, \(\frac{di}{dt} = 100 \, \text{A/s}\). We will use the formula for back emf induced in a coil.
02

Recall the Formula for Back EMF

The back emf (\(\epsilon\)) induced in an inductor is given by the formula \(\epsilon = -L \frac{di}{dt}\). This equation states that the induced emf is proportional to the rate of change of current through the coil and the inductance of the coil.
03

Substitute Given Values

Substitute the given values into the formula: \( \epsilon = -8.20 \, \text{mH} \times 100 \, \text{A/s} \). Convert millihenrys to henrys, where \(1 \, \text{mH} = 0.001 \, \text{H}\), thus \(8.20 \, \text{mH} = 0.0082 \, \text{H}\).
04

Calculate the Back EMF

Perform the multiplication: \( \epsilon = -0.0082 \, \text{H} \times 100 \, \text{A/s} = -0.82 \, \text{V}\). Note the negative sign indicates the direction of the induced emf is opposite to the direction of the change in current.

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Key Concepts

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

Self-Inductance
Self-inductance is a fundamental property of a coil or an inductor. It refers to the ability of the coil to induce an electromotive force (EMF) in itself, whenever there is a change in the current flowing through it. This property is described by the symbol 'L'.
It's primarily measured in henrys (H) and indicates how effectively the coil can oppose any change in the current due to its own magnetic field. Essentially, self-inductance can be thought as the 'inertia' against changes in current.
Here's how it works:
  • When current flows through a coil, it generates a magnetic field around it.
  • If this current changes, the resulting change in the magnetic field will, in turn, induce a voltage in the same coil, either adding to or opposing the change.
The larger the value of self-inductance, the greater the induced voltage for a given rate of change in current. This is why coils with higher inductance are more effective at opposing changes in current.
Rate of Change of Current
The rate of change of current is an important concept when studying inductance. It refers to how quickly the current is increasing or decreasing through a coil.Mathematically, this rate is represented as \( \frac{di}{dt} \), where \(i\) is the current, and \(t\) is time. The unit of this rate is amperes per second (A/s), describing how many amperes the current changes each second.
This rate plays a crucial role in determining the back emf generated in the coil:
  • The faster the current changes, the greater the back emf will be for a given self-inductance.
  • Rate of change could either be positive (increasing current) or negative (decreasing current).
For practical purposes, managing this rate becomes crucial in various electronic devices and applications to prevent potentially damaging back emf.
Induced EMF Formula
The induced electromotive force (EMF) in a coil can be calculated using a straightforward formula based on the concepts of self-inductance and the rate of change of current. This formula is expressed as \( \epsilon = -L \frac{di}{dt} \).
Let's break it down:
  • \( \epsilon \) represents the induced emf, measured in volts (V).
  • \(L\) denotes the self-inductance of the coil, in henrys (H).
  • \( \frac{di}{dt} \) is the rate of change of current, in amperes per second (A/s).
The negative sign in the formula illustrates Lenz's Law, indicating that the induced emf will always act to oppose the change in the current that produced it.
Lenz's Law is crucial in all applications involving electromagnetism, ensuring that the induced emf does not simply increase indefinitely, which could disrupt the flow of current in circuits if not properly accounted for.

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Most popular questions from this chapter

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?

An emf of \(8.0 \mathrm{~V}\) is induced in a coil when the current in it changes at the rate of \(32 \mathrm{~A} / \mathrm{s}\). Compute the inductance of the coil.

When the current in a certain coil is changing at a rate of \(3.0 \mathrm{~A} / \mathrm{s}\), it is found that an emf of \(7.0 \mathrm{mV}\) is induced in a nearby coil. What is the mutual inductance of the combination? $$ \ell_{s}=M \frac{\Delta i_{p}}{\Delta t} \text { or } M=\varepsilon_{s} \frac{\Delta t}{\Delta i_{p}}=\left(7.0 \times 10^{-3} \mathrm{~V}\right) \frac{1.0 \mathrm{~s}}{3.0 \mathrm{~A}}=2.3 \mathrm{mH} $$

A certain series circuit consists of a 12-V battery, a switch, a 1.0\(\mathrm{M} \Omega\) resistor, and a \(2.0-\mu \mathrm{F}\) capacitor, initially uncharged. If the switch is now closed, find \((a)\) the initial current in the circuit, \((b)\) the time for the current to drop to \(0.37\) of its initial value, \((c)\) the charge on the capacitor then, and \((d)\) the final charge on the capacitor. (a) The loop rule applied to the circuit of Fig. \(34-1(a)\) at any instant gives $$ 12 \mathrm{~V}-i R-u_{c}=0 $$ where \(v_{c}\) is the p.d. across the capacitor. At the first instant, \(q\) is essentially zero and so \(u_{c}=0\). Then $$ 12 \mathrm{~V}-i R-0=0 \quad \text { or } \quad i=\frac{12 \mathrm{~V}}{1.0 \times 10^{6} \Omega}=12 \mu \mathrm{A} $$ (b) The current drops to \(0.37\) of its initial value when $$ t=R C=\left(1.0 \times 10^{6} \Omega\right)\left(2.0 \times 10^{-6} \mathrm{~F}\right)=2.0 \mathrm{~s} $$ (c) At \(t=2.0 \mathrm{~s}\) the charge on the capacitor has increased to \(0.63\) of its final value. [See part ( \(d\) ) below.] (d) The charge ceases to increase when \(i=0\) and \(v_{c}=12 \mathrm{~V}\). Therefore, $$ q_{\text {final }}=C u c=\left(2.0 \times 10^{-6} \mathrm{~F}\right)(12 \mathrm{~V})=24 \mu \mathrm{C} $$

I A charged capacitor is connected across a \(10-\mathrm{k} \Omega\) resistor and allowed to discharge. The potential difference across the capacitor drops to \(0.37\) of its original value after a time of \(7.0 \mathrm{~s}\). What is the capacitance of the capacitor?
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