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A toroidal inductor with an inductance of \(90.0 \mathrm{mH}\) encloses a volume of \(0.0200 \mathrm{~m}^{3}\). If the average energy density in the toroid is \(70.0 \mathrm{~J} / \mathrm{m}^{3}\), what is the current through the inductor?

Short Answer

Expert verified
The current through the inductor is approximately 5.58 A.

Step by step solution

01

Understand the Relationship Between Energy Density and Inductor Parameters

The energy density in an inductor is the energy stored per unit volume, given by \( u = \frac{1}{2} B^2 / \mu_0 \) for a magnetic field \( B \). For an inductor, the energy \( U \) is also given by \( U = \frac{1}{2} L I^2 \), where \( L \) is the inductance and \( I \) is the current.
02

Calculate Total Energy Stored in the Toroid

The total energy stored in the toroidal inductor can be found from the average energy density and the volume it encloses. Use the formula: \[ U = u \times \text{Volume} = 70.0 \frac{\mathrm{J}}{\mathrm{m}^3} \times 0.0200 \mathrm{~m}^3 \]This results in:\[ U = 1.4 \mathrm{~J} \]
03

Relate Total Energy to Current

Using the relationship between energy stored and the inductance, substitute the given energy to solve for current:\[ U = \frac{1}{2} L I^2 \]Substituting \( U = 1.4 \mathrm{~J} \) and \( L = 90.0 \mathrm{mH} = 0.090 \mathrm{~H} \):\[ 1.4 = \frac{1}{2} \cdot 0.090 \cdot I^2 \]
04

Solve for Current

Rearrange the equation from the previous step to solve for the current \( I \):\[ I^2 = \frac{2 \times 1.4}{0.090} \]\[ I^2 = \frac{2.8}{0.090} = 31.11 \]Take the square root to find \( I \):\[ I = \sqrt{31.11} \approx 5.58 \mathrm{A} \]

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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 that describes how effectively an electrical conductor, such as a coil, can store energy when a current flows through it. Its unit is the Henry (H). Inductance in a circuit can be thought of as the "magnetic inertia," which resists changes in the current passing through it. The more turns a coil has, the higher its inductance will be, because more turns create more magnetic fields which oppose the change in current.

The inductance equation is given by \( L \), which in the exercise, has a value of \(90.0 \mathrm{mH}\) (millihenrys) or \(0.090 \mathrm{~H}\) when converted. When we refer to inductance in devices like toroidal inductors, we are often looking at how they store energy in a magnetic field. This storage capability is crucial in applications like transformers and inductive sensors, where control of magnetic fields is needed.
  • Inductance opposes sudden changes in current flow.
  • High inductance means more energy storage in the magnetic field.
  • Inductive components are key in filtering and signal processing circuits.
Energy Density
Energy density refers to the amount of energy stored in a given system or region of space per unit volume. In electromagnetism, energy density is particularly useful in describing the distribution of energy when it is stored in devices like inductors. For an inductor, energy density \( u \) can be calculated using the formula:\[ u = \frac{B^2}{2\mu_0} \]where \( B \) is the magnetic field and \( \mu_0 \) is the permeability of free space.

In the toroidal inductor exercise, the given energy density is \(70.0 \mathrm{~J} / \mathrm{m}^3\). This means that for every cubic meter of space inside the inductor, there are 70 Joules of energy stored. Calculating energy density allows engineers to understand how effectively a device can hold energy, which is essential for efficient design and optimization.
  • Energy density indicates how densely energy is packed in a space.
  • Higher energy density means more energy can be stored in the same volume.
  • It is crucial for developing compact and efficient energy storage solutions.
Toroidal Inductor
A toroidal inductor is a type of inductor where the coil is shaped like a torus, or a doughnut. This shape is highly efficient for magnetic field containment, as it confines the magnetic field within the core material, reducing electromagnetic interference with surrounding circuitry. The efficiency makes the toroidal design highly desirable in electronic applications such as power supply units and audio equipment.

In the problem, the toroidal inductor encloses a volume of \(0.0200 \mathrm{~m}^3\). The volume is crucial because it indicates how much space the magnetic field occupies and directly relates to the energy density calculations for the system.
  • Toroidal inductors minimize the loss of magnetic field energy.
  • The shape ensures that most of the magnetic field lines remain inside the core.
  • They are compact and efficiently store energy, making them ideal for various electrical systems.

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Most popular questions from this chapter

At time \(t=0\), a \(45 \mathrm{~V}\) potential difference is suddenly applied to the leads of a coil with inductance \(L=50 \mathrm{mH}\) and resistance \(R=180 \Omega\). At what rate is the current through the coil increasing at \(t=1.2 \mathrm{~ms}\) ?

A coil with an inductance of \(2.0 \mathrm{H}\) and a resistance of \(10 \Omega\) is suddenly connected to an ideal battery with \(\mathscr{8}=100 \mathrm{~V} .\) At \(0.10 \mathrm{~s}\) after the connection is made, what is the rate at which (a) energy is being stored in the magnetic field, (b) thermal energy is appearing in the resistance, and (c) energy is being delivered by the battery?

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?

A long solenoid has a diameter of \(12.0 \mathrm{~cm}\). When a current \(i\) exists in its windings, a uniform magnetic field of magnitude \(B=30.0 \mathrm{mT}\) is produced in its interior. By decreasing \(i\), the field is caused to decrease at the rate of \(6.50 \mathrm{mT} / \mathrm{s}\). Calculate the magnitude of the induced electric field (a) \(2.20 \mathrm{~cm}\) and (b) \(8.20 \mathrm{~cm}\) from the axis of the solenoid.

At a certain place, Earth's magnetic field has magnitude \(B=0.590\) gauss and is inclined downward at an angle of \(70.0^{\circ}\) to the horizontal. A flat horizontal circular coil of wire with a radius of \(10.0 \mathrm{~cm}\) has 1000 turns and a total resistance of \(85.0 \Omega\). It is connected in series to a meter with \(140 \Omega\) resistance. The coil is flipped through a half- revolution about a diameter, so that it is again horizontal. How much charge flows through the meter during the flip?

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