/*! 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 15 An emf of \(8.0 \mathrm{~V}\) is... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The inductance of the coil is \(0.25 \mathrm{~H}\).

Step by step solution

01

Understand Faraday's Law of Induction

Faraday's Law of Induction states that the induced electromotive force (emf) in a coil is directly proportional to the rate of change of current through the coil and the self-inductance of the coil. Mathematically, this is expressed as: \( \text{emf} = -L \frac{di}{dt} \), where \( L \) is the inductance, \( \text{emf} \) is the induced emf, and \( \frac{di}{dt} \) is the rate of change of current.
02

Determine Given Values

From the problem, we know the following: \( \text{emf} = 8.0 \mathrm{~V} \) and \( \frac{di}{dt} = 32 \mathrm{~A/s} \). You need to find the inductance \( L \) of the coil.
03

Rearrange the Equation

We want to find \( L \) in the equation \( \text{emf} = -L \frac{di}{dt} \). Rearranging the equation gives us \( L = -\frac{\text{emf}}{\frac{di}{dt}} \). Note that the negative sign indicates direction and can be ignored when calculating magnitude.
04

Substitute the Values

Substitute the given values into the rearranged equation: \( L = \frac{8.0}{32} \).
05

Calculate the Inductance

Perform the division to get the inductance of the coil: \( L = 0.25 \mathrm{~H} \).

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Key Concepts

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

Faraday's Law of Induction
Faraday's Law of Induction is a fundamental principle describing how electric currents can be induced in a coil or loop by changing magnetic fields. This effect was discovered by Michael Faraday in the 19th century, and it is a cornerstone of electromagnetism. The law states that the electromotive force (emf) induced in a coil is proportional to the rate at which the magnetic field through the coil is changing. This relationship is mathematically expressed as:\[ \text{emf} = -L \frac{di}{dt} \]where:
  • emf is the induced electromotive force in volts (V),
  • L is the inductance of the coil in henrys (H),
  • \(\frac{di}{dt}\) is the rate of change of current in amperes per second (A/s), and
  • the negative sign indicates the direction of the induced emf, as per Lenz's Law, which states that the induced current will oppose the change in magnetic flux.
This means that whenever there is a change in the magnetic field around a conductor, an electric current is generated. This is how electric generators work.
Electromotive Force (emf)
Electromotive force, or emf, is a term used to describe the energy provided by a source to move electrons through a circuit. It is measured in volts, just like potential difference. However, emf specifically refers to the energy supplied to generate an electric current, not the energy used once the current is flowing. In a coil, the emf is generated whenever there is a change in the magnetic environment of the coil.
It's important to remember that emf is not actually a "force" but rather a potential difference. It is commonly studied in systems where magnetic fields change, such as transformers, electric motors, and inductors in circuits.
  • For instance, in our problem, the emf is given as 8.0 V. This is the potential difference that was induced in the coil.
  • It's this change in potential difference that then drives a current through the circuit, resulting in electromagnetic induction.
Understanding emf is crucial for mastering how electric circuits respond to changing currents and magnetic fields.
Rate of Change of Current
The rate of change of current, denoted \( \frac{di}{dt} \), represents how quickly the current in a circuit is changing with time. It is measured in amperes per second (A/s). This concept is fundamental in determining the behavior of inductors in an electrical circuit.
For coils or inductors, the change of current over time directly affects how much emf is induced across the coil.
  • In our example problem, the rate of change of current is given as 32 A/s.
  • This value tells us how quickly the current is increasing or decreasing, which influences the magnitude of the induced emf.
Calculating \( \frac{di}{dt} \) becomes especially important in alternating current (AC) circuits where currents are continuously changing. This calculation helps engineers and scientists design circuits that handle or exploit these changes efficiently. Ultimately, the rate of change of current is a key factor in how circuits respond dynamically.
Coil Inductance
Coil inductance is a property of electrical circuits that quantifies how well they can support the creation of induced emf when exposed to changing currents. It is measured in henrys (H), named after Joseph Henry, an American scientist who discovered electromagnetic induction independently of Michael Faraday. The inductance of a coil depends on several factors:
  • The number of turns in the coil – more turns increase inductance.
  • The coil's size and shape – larger coils with larger cross-sectional areas have greater inductance.
  • The core material – coils with a ferromagnetic core have greater inductance.
For our problem, the inductance value calculated was 0.25 H. This value tells us how well the coil can induce voltage in itself as the current changes. In practical applications, inductors are used in various electrical devices to manage signal processing, filter signals, and maintain stable currents. Understanding inductance is vital for anyone designing circuits that involve electromagnetism and energy transfer.

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Most popular questions from this chapter

A \(2.0-\mu \mathrm{F}\) capacitor is charged through a \(30-\mathrm{M} \Omega\) resistor by a 45-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, $$ q_{\infty}=V_{\infty} C=(45 \mathrm{~V})\left(2.0 \times 10^{-6} \mathrm{~F}\right)=9.0 \times 10^{-5} \mathrm{C} $$ (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 \(e^{-83 / 60}=e^{-1.383}=0.25\) 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 steady current of \(2.5\) A creates a flux of \(1.4 \times 10^{-4} \mathrm{~Wb}\) in a coil of 500 turns. What is the inductance of the coil?

Two coils are wound on the same iron rod so that the flux generated by one passes through the other also. The primary coil has \(N_{p}\) loops and, when a current of \(2.0 \mathrm{~A}\) flows through it, the flux in it is \(2.5 \times 10^{-4} \mathrm{~Wb}\). Determine the mutual inductance of the two coils if the secondary coil has \(N_{s}\) loops. $$ \left|\delta_{s}\right|=N_{s}\left|\frac{\Delta \Phi_{M s}}{\Delta t}\right| \text { and }\left|\delta_{s}\right|=M\left|\frac{\Delta i_{p}}{\Delta t}\right| $$ give \(\quad M=N_{s}\left|\frac{\Delta \Phi_{M s}}{\Delta i_{p}}\right|=N_{s} \frac{\left(2.5 \times 10^{-4}-0\right) \mathrm{Wb}}{(2.0-0) \mathrm{A}}=\left(1.3 \times 10^{-4} \mathrm{~N}_{s}\right) \mathrm{H}\)

The mutual inductance between the primary and secondary of a transformer is \(0.30 \mathrm{H}\). Compute the induced emf in the secondary when the primary current changes at the rate of \(4.0 \mathrm{~A} / \mathrm{s}\).

At a certain instant, a coil with a resistance of \(0.40 \Omega\) and a self- inductance of \(200 \mathrm{mH}\) carries a current of \(0.30 \mathrm{~A}\) that is increasing at the rate of \(0.50 \mathrm{~A} / \mathrm{s} .(a)\) What is the potential difference across the coil at that instant? \((b)\) Repeat if the current is decreasing at \(0.50 \mathrm{~A} / \mathrm{s}\). We can represent the coil by a resistance in series with an emf (the induced emf), as shown in Fig. 34.3. (a) Because the current is increasing, \(\mathscr{E}\) will oppose the current and therefore have the polarity shown. We write the loop equation for the circuit: $$ V_{b a}-i R-\varepsilon=0 $$ Since \(V_{b a}\) is the voltage across the coil, and since \(\mathscr{E}=L|\Delta i / \Delta t|\), we have $$ V_{\text {coil }}=i R+\ell=(0.30 \mathrm{~A})(0.40 \Omega)+(0.200 \mathrm{H})(0.50 \mathrm{~A} / \mathrm{s})=0.22 \mathrm{~V} $$ (b) With \(i\) decreasing, the induced emf must be reversed in Fig. \(34-3 .\) This gives \(V_{\text {coil }}=i R-\varepsilon=0.020 \mathrm{~V}\).

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