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Chapter 33: Problem 7

Energy Density A solenoid that is \(85.0 \mathrm{~cm}\) long has a cross- sectional area of \(17.0 \mathrm{~cm}^{2}\). There are 950 turns of wire carrying a current of \(6.60\) A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Short Answer

Expert verified
The energy density is \( u = \frac{B^2}{2\text{\textmu}_0} \). The total energy stored is \( U = u \times V \).

Step by step solution

01

Calculate the Magnetic Field Inside the Solenoid

The formula for the magnetic field inside a solenoid is given by \[ B = \frac{\text{\textmu}_0 \times N \times I}{L} \]where N = 950 \text{ (number of turns)},I = 6.60 \text{ A (current)},L = 85.0 \text{ cm} = 0.850 \text{ m (length)},\(\text{\textmu}_0 = 4\text{\textpi} \times 10^{-7} \text{ T}\text{m/A}\) (permeability of free space). Substitute the values to get:\[ B = \frac{4\text{\textpi} \times 10^{-7} \times 950 \times 6.60}{0.850} \]Calculate \(B\) to find the magnetic field inside the solenoid.
02

Calculate the Energy Density

The energy density (u) of the magnetic field is given by the formula \[ u = \frac{B^{2}}{2\text{\textmu}_0} \]Use the magnetic field (B) calculated in Step 1 and substitute it into the formula to find u.
03

Identify Parameters for Total Energy Calculation

The total energy (U) stored in the magnetic field can be found using the formula\[ U = u \times V \]where u is the energy density from Step 2 and V is the volume of the solenoid. The volume V is calculated by multiplying the cross-sectional area (A = 17.0 cm^2 = 17.0 \times 10^{-4} m^2) by the length (L = 0.850 m).
04

Calculate Volume of Solenoid

Calculate the volume V of the solenoid using\[ V = A \times L \]Substitute the given values to find V:\[ V = 17.0 \times 10^{-4} \text{ m}^2 \times 0.850 \text{ m} \]
05

Calculate Total Energy Stored

Using the energy density u from Step 2 and the volume V from Step 4, calculate the total energy stored using\[ U = u \times V \]

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Key Concepts

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

Energy Density
The energy density of a magnetic field represents the amount of energy stored per unit volume in the field. It's given by the formula: \[ u = \frac{B^{2}}{2\text{\textmu}_0} \] Here, \( B \) is the magnetic field strength, and \( \text{\textmu}_0 \) is the permeability of free space, which is \( 4\text{\textpi} \times 10^{-7} \text{ T} \text{ m/A} \). Knowing this, to find the energy density inside the solenoid, follow these steps:
  • First, calculate the magnetic field inside the solenoid using the number of turns (N), the current (I), and the length (L) of the solenoid.
  • Then, plug the calculated magnetic field (B) value into the energy density formula.
So, once you determine \( B \), you simply square it, divide by twice the permeability of free space, and you get the energy density.
Solenoid Calculations
Calculating parameters for a solenoid involves a few key steps. A solenoid is essentially a coil of wire designed to produce a magnetic field when an electric current passes through it. Here are the steps to solve for the magnetic field and volume of the solenoid:
  • First, calculate the Magnetic Field (B): \[ B = \frac{\text{\textmu}_0 \times N \times I}{L} \] Plug in the given values (number of turns N = 950, current I = 6.60 A, length L = 0.850 m) and use the value for \textmu_0\ (\( 4\text{\textpi} \times 10^{-7} \text{ T}\text{ m/A} \)).
  • To get the volume (V) of the solenoid, you need the cross-sectional area (A) and its length (L):\[ V = A \times L \] The given cross-sectional area should be converted to meters squared (A = 0.0017 m\text{\textsuperscript{2}}) and then multiplied by the length (L = 0.85 m).
Following these steps will give you the essential parameters needed for further calculations, such as energy density and total energy stored.
Magnetic Field Energy
The total energy stored in the magnetic field within the solenoid can be determined using the energy density and the volume of the solenoid. The formula for total energy stored (U) is: \[ U = u \times V \] Here, \( u \) is the energy density calculated from the magnetic field, and \( V \) is the volume of the solenoid.Steps to determine the total energy stored:
  • First, find the energy density \( u \) using the previously calculated magnetic field (B).
  • Next, calculate the volume \( V \) of the solenoid using its cross-sectional area and length.
  • Finally, multiply the energy density by the solenoid's volume to get the total energy stored.
This approach allows us to understand and quantify the magnetic field's potential to store energy in a given space. It's crucial for applications where magnetic fields are used to store and transfer energy efficiently.

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Most popular questions from this chapter

SHM A \(0.50 \mathrm{~kg}\) body oscillates in simple harmonic motion on a spring that, when extended \(2.0 \mathrm{~mm}\) from its equilibrium, has an \(8.0 \mathrm{~N}\) restoring force. (a) What is the angular frequency of oscillation? (b) What is the period of oscillation? (c) What is the capacitance of an \(L C\) circuit with the same period if \(L\) is chosen to be \(5.0 \mathrm{H}\) ?

Energy Is Dissipated In an \(R L C\) circuit such as that of assume that \(R=5.00 \Omega, L=60.0 \mathrm{mH} f^{\mathrm{dr}}=60.0 \mathrm{~Hz}\), and \(\left|\mathscr{E}^{\max }\right|=30.0 \mathrm{~V}\). For what values of the capacitor would the average rate at which energy is dissipated in the resistance be (a) a maximum and (b) a minimum? (c) What are these maximum and minimum energy dissipation rates? What are (d) the corresponding phase angles and (e) the corresponding power factors?

Unknown Resistance A coil of inductance \(88 \mathrm{mH}\) and unknown resistance and a \(0.94 \mu \mathrm{F}\) capacitor are connected in series with an alternating emf of frequency \(930 \mathrm{~Hz}\). If the phase constant between the applied voltage and the current is \(75^{\circ}\), what is the resistance of the coil?

Maximum Charge In an oscillating \(L C\) circuit, \(L=1.10 \mathrm{mH}\) and \(C=4.00 \mu \mathrm{F}\). The maximum charge on the capacitor is \(3.00 \mu \mathrm{C}\). Find the maximum current.

Electric to Magnetic Energy In a certain oscillating \(L C\) circuit the total energy is converted from electric energy in the capacitor to magnetic energy in the inductor in \(1.50 \mu \mathrm{s}\). (a) What is the period of oscillation? (b) What is the frequency of oscillation? (c) How long after the magnetic energy is a maximum will it be a maximum again?
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