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A coil with an inductance of \(2.0 \mathrm{H}\) and a resistance of \(10 \Omega\) is suddenly connected to an ideal battery with \(\mathscr{B}=100 \mathrm{~V} .\) (a) What is the equilibrium current? (b) How much energy is stored in the magnetic field when this current exists in the coil?

Short Answer

Expert verified
(a) Equilibrium current is 10 A; (b) Energy stored is 100 J.

Step by step solution

01

Calculate Equilibrium Current

When the coil is connected to the battery and the current reaches equilibrium, the inductive effects disappear, and only the resistance affects the circuit. By Ohm's Law, we have:\[ I = \frac{\mathscr{B}}{R} \]where \( \mathscr{B} = 100 \mathrm{~V} \) and \( R = 10 \Omega \). Thus:\[ I = \frac{100 \mathrm{~V}}{10 \Omega} = 10 \mathrm{~A} \].
02

Calculate Energy Stored in the Magnetic Field

The energy stored in the magnetic field in an inductor is given by the formula:\[ U = \frac{1}{2}LI^2 \]Using \( L = 2.0 \mathrm{~H} \) and our previously calculated equilibrium current \( I = 10 \mathrm{~A} \), we find:\[ U = \frac{1}{2} \times 2.0 \mathrm{~H} \times (10 \mathrm{~A})^2 \]\[ U = \frac{1}{2} \times 2.0 \times 100 = 100 \mathrm{~J} \].

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Key Concepts

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

Inductance
Inductance is a fundamental concept in electromagnetism representing the ability of a coil to store energy in a magnetic field due to current flowing through it. Think of it as the "inertia" of an electrical circuit. When the current changes, the inductor resists these changes by inducing a voltage that opposes them. This is a first principle of Lenz's Law. Inductance is measured in Henrys (H).

In a practical example, consider our coil with an inductance of 2.0 H. Such a coil can significantly oppose changes in current, meaning that the energy isn't immediately consumed but stored for later use. The inductance determines how much energy can be stored, among other things, by the formula for magnetic field energy. However, remember that the inductor only opposes changes and does not affect a steady current.

When the circuit reaches equilibrium, and the inductive effects disappear, we focus only on the resistance of the circuit for further calculations.
Ohm's Law
Ohm's Law is a fundamental rule in electrical circuits. It expresses a simple relationship between voltage (\( V \)), current (\( I \)), and resistance (\( R \)). It is given by:
  • \( I = \frac{V}{R} \)
For our exercise, applying Ohm's Law helps calculate the equilibrium current when the coil is connected to a battery. Using a battery voltage (\( \mathscr{B} \)) of 100 V and resistance (\( R \)) of 10 Ω, the current (\( I \)) is calculated as:
  • \( I = \frac{100 \mathrm{~V}}{10 \Omega} = 10 \mathrm{~A} \)
Here, resistance is the only factor determining the current since the inductive effects have faded at equilibrium.

Ohm's Law is crucial not just for calculations but also for understanding how voltage, current, and resistance interact in any electrical circuit. It's a tool that simplifies measuring and designing circuits by predicting how they will behave under various conditions.
Magnetic Field Energy
magnetic field energy is the energy stored in a coil or inductor when current passes through it. This energy comes from the magnetic field created around the inductor as the electric current flows. The amount of energy depends on both the inductance and the square of the current.

The formula describing this energy is:
  • \( U = \frac{1}{2}LI^2 \)
Where \( U \) is the energy stored, \( L \) is the inductance, and \( I \) is the current. For the coil in our original exercise with \( L = 2.0 \mathrm{H} \) and an equilibrium current \( I = 10 \mathrm{~A} \), it helps us calculate the magnetic field energy as:
  • \( U = \frac{1}{2} \times 2.0 \mathrm{~H} \times (10 \mathrm{~A})^2 = 100 \mathrm{~J} \)
Understanding how energy is stored and released in magnetic fields is essential for designing circuits, especially in devices that rely on inductors for energy storage. This stored energy can be significant and finds practical applications in transformers, inductors, and many electrical devices.

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Most popular questions from this chapter

An electric generator contains a coil of 100 turns of wire, each forming a rectangular loop \(50.0 \mathrm{~cm}\) by \(30.0 \mathrm{~cm} .\) The coil is placed entirely in a uniform magnetic field with magnitude \(B=\) \(3.50 \mathrm{~T}\) and with \(\vec{B}\) initially perpendicular to the coil's plane. What is the maximum value of the emf produced when the coil is spun at 1000 rev/min about an axis perpendicular to \(\vec{B}\) ?

If \(50.0 \mathrm{~cm}\) of copper wire (diameter \(=1.00 \mathrm{~mm}\) ) is formed into a circular loop and placed perpendicular to a uniform magnetic field that is increasing at the constant rate of \(10.0 \mathrm{mT} / \mathrm{s}\), at what rate is thermal energy generated in the loop?

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a \(0.50 \mathrm{~T}\) magnetic field?

Two inductors \(L_{1}\) and \(L_{2}\) are connected in series and are separated by a large distance so that the magnetic field of one cannot affect the other. (a) Show that the equivalent inductance is given by $$ L_{\mathrm{eq}}=L_{1}+L_{2} $$ (Hint: Review the derivations for resistors in series and capacitors in series. Which is similar here?) (b) What is the generalization of (a) for \(N\) inductors in series?

The inductance of a closely wound coil is such that an emf of \(3.00 \mathrm{mV}\) is induced when the current changes at the rate of \(5.00\) A/s. A steady current of \(8.00\) A produces a magnetic flux of \(40.0\) \(\mu \mathrm{Wb}\) through each turn. (a) Calculate the inductance of the coil. (b) How many turns does the coil have?

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