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It is proposed to store \(1.00 \mathrm{kW} \cdot \mathrm{h}=3.60 \times 10^{6} \mathrm{J}\) of electrical energy in a uniform magnetic field with magnitude 0.600 \(\mathrm{T}\) . (a) What volume (in vacuum) must the magnetic field occupy to store this amount of energy? (b) If instead this amount of energy is to be stored in a volume (in vacuum) equivalent to a cube 40.0 \(\mathrm{cm}\) on a side, what magnetic field is required?

Short Answer

Expert verified
(a) 25.12 m³; (b) 5.94 T

Step by step solution

01

Understanding Magnetic Energy Density

From physics, the energy density (energy per unit volume) of a magnetic field can be expressed as \( u_B = \frac{B^2}{2\mu_0} \), where \( B \) is the magnetic field strength and \( \mu_0 \) is the permeability of free space (\( \mu_0 = 4\pi \times 10^{-7} \mathrm{T}\cdot\mathrm{m/A} \)).
02

Calculate Energy Density

First, we need to compute the energy density of the magnetic field for the given magnetic field strength of 0.600 T. Substitute \( B = 0.600 \mathrm{T} \) into the formula:\[ u_B = \frac{(0.600)^2}{2 \times 4\pi \times 10^{-7}} = \frac{0.36}{2.513 \times 10^{-6}} \approx 1.433 \times 10^5 \text{ J/m}^3 \].
03

Calculate Required Volume

The total energy to be stored is \( 3.60 \times 10^6 \mathrm{J} \). To find the volume \( V \) needed, use the formula \( U = u_B \times V \):\[ V = \frac{3.60 \times 10^6}{1.433 \times 10^5} \approx 25.12 \text{ m}^3 \].
04

Determine Volume of a Cube

Given a cube with each side of 40.0 cm, the volume \( V \) is calculated as follows:\[ V = (0.40)^3 = 0.064 \text{ m}^3 \].
05

Find Required Magnetic Field for Cube

To store the same amount of energy \( 3.60 \times 10^6 \mathrm{J} \) in 0.064 \text{ m}^3, use the energy density and rearrange to find \( B \):\[ u_B = \frac{3.60 \times 10^6}{0.064} = 5.625 \times 10^7 \text{ J/m}^3 \].Solving for \( B \):\[ \frac{B^2}{2 \times 4\pi \times 10^{-7}} = 5.625 \times 10^7 \]\[ B^2 = 2 \times 4\pi \times 10^{-7} \times 5.625 \times 10^7 \approx 3.53 \times 10^1 \]\[ B \approx \sqrt{35.3} \approx 5.94 \mathrm{T} \].

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Key Concepts

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

Energy Density
Energy density is a core concept when considering the storage of energy in magnetic fields. It refers to the amount of energy stored in a given volume. For magnetic fields, energy density (\( u_B \)) can be calculated using the formula:
  • \( u_B = \frac{B^2}{2\mu_0} \)
Here, \( B \) is the magnetic field strength, and \( \mu_0 \) is the permeability of free space. This equation helps you understand how much energy can be stored for a particular magnetic field strength. It shows there's a squared relationship between energy density and magnetic field strength. This squared dependency highlights that even small increases in the magnetic field dramatically increase the energy stored. Knowing the energy density is crucial for calculating the volume required to store a given amount of energy.
Magnetic Field Strength
Magnetic field strength, denoted by \( B \), is a crucial factor in determining how much energy can be stored in a magnetic field. It's measured in Tesla (T) and reflects the field's intensity. A higher magnetic field strength means a denser energy storage for a given volume, as seen in the energy density formula:
  • \( u_B = \frac{B^2}{2\mu_0} \)
It is important to note that increasing \( B \) will lead to a more significant energy storage capacity, exponentially growing with its square. This property is particularly useful when space is a constraint, allowing for high energy storage in compact volumes by enhancing the field strength – assuming materials can withstand such intense fields.
Magnetic Field Energy
Magnetic field energy refers to the total energy stored within a magnetic field over a certain volume. Calculating the total magnetic energy involves multiplying the energy density by the volume occupied by the magnetic field:
  • \( U = u_B \times V \)
This expression highlights two critical influences on magnetic field energy: the energy density (which depends on the magnetic field strength, \( B \)) and the volume \( V \) the field occupies. To tailor energy storage solutions, an understanding of this relationship allows for adjustments in either the magnetic field strength or the storage volume, based on what resource is readily available or limited. This principle is utilized when designing systems like superconducting magnetic energy storage (SMES) devices.
Permeability of Free Space
The concept of permeability of free space, also known as the magnetic constant, is denoted by \( \mu_0 \). It is a critical constant in physics, equal to \( 4\pi \times 10^{-7} \mathrm{T}\cdot\mathrm{m/A} \).
This constant appears prominently in formulas related to magnetic fields, such as the energy density formula \( u_B = \frac{B^2}{2\mu_0} \). The permeability of free space relates the magnetic field to the generated field lines within a vacuum. It signifies the ability of a vacuum to support a magnetic field. As a constant, \( \mu_0 \) ensures standardization across calculations, facilitating accurate predictions and comparisons. Understanding this property is essential for quantifying and predicting magnetic phenomena, especially in applications like magnetic energy storage.

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Most popular questions from this chapter

An inductor is connected to the terminals of a battery that has an emf of \(12.0 \mathrm{~V}\) and negligible internal resistance. The current is \(4.86 \mathrm{~mA}\) at \(0.725 \mathrm{~ms}\) after the connection is completed. After a long time the current is \(6.45 \mathrm{~mA}\). What are (a) the resistance \(R\) of the inductor and (b) the inductance \(L\) of the inductor?

Two toroidal solenoids are wound around the same form so that the magnetic field of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is 6.52 A, the average flux through each turn of solenoid 2 is 0.0320 Wb. (a) What is the mutual inductance of the pair of solenoids? (b) When the current in solenoid 2 is 2.54 \(\mathrm{A}\) , what is the average flux through each turn of solenoid 1\(?\)

L-C Oscillations. A capacitor with capacitance \(6.00 \times\) \(10^{-5} \mathrm{F}\) is charged by connecting it to a \(12.0-\mathrm{V}\) battery. The capacitor is disconnected from the battery and connected across an inductor with \(L=1.50 \mathrm{H}\) (a) What are the angular frequency \(\omega\) of the electrical oscillations and the period of these oscillations (the time for one oscillation \() ?(\text { b) What is the initial charge on the }\) capacitor? (c) How much energy is initially stored in the capacitor? (d) What is the charge on the capacitor 0.0230 \(\mathrm{s}\) after the connection to the inductor is made? Interpret the sign of your answer.(e) At the time given in part (d), what is the current in the inductor? Interpret the sign of your answer. (f) At the time given in part (d), how much electrical energy is stored in the capacitor and how much is stored in the inductor?

A 20.0 -\muF capacitor is charged by a \(150.0-\mathrm{V}\) power supply, then disconnected from the power and connected in series with a \(0.280-\mathrm{mH}\) inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time \(t=0 \mathrm{ms}\) (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t=1.30 \mathrm{ms}\) .

A solenoid 25.0 \(\mathrm{cm}\) long and with a cross-sectional area of 0.500 \(\mathrm{cm}^{2}\) contains 400 turns of wire and carries a current of 80.0 A. Calculate: (a) the magnetic field in the solenoid; (b) the energy density in the magnetic fleld if the solenoid is filled with air; (c) the total energy contained in the coil's magnetic field (assume the field is uniform); (d) the inductance of the solenoid.

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