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Chapter 30: Problem 19

A solenoid 25.0 cm long and with a cross-sectional area of 0.500 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 field 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.

Short Answer

Expert verified
(a) \( B = 0.403 \text{ T}\); (b) \( u = 645 \text{ J/m}^3 \); (c) \( U = 0.0806 \text{ J}\); (d) \( L_s = 0.0101 \text{ H}\).

Step by step solution

01

Calculate Magnetic Field Inside the Solenoid

The magnetic field inside a solenoid is given by the formula \( B = \mu_0 \frac{N}{L} I \), where \( \mu_0 \) is the permeability of free space \( (\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}) \), \( N \) is the number of turns, \( L \) is the length, and \( I \) is the current. Substituting the values given: \( N = 400 \), \( L = 0.25 \text{ m} \), \( I = 80.0 \text{ A} \):\[ B = 4\pi \times 10^{-7} \times \frac{400}{0.25} \times 80 \]Calculate to find the magnetic field \( B \) inside the solenoid.
02

Calculate the Energy Density in the Magnetic Field

The energy density \( u \) in the magnetic field is given by \( u = \frac{B^2}{2\mu_0} \). Use the magnetic field \( B \) calculated in Step 1 and \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \) :\[ u = \frac{(B^2)}{2 \times 4\pi \times 10^{-7}} \]Calculate \( u \) using the value of \( B \).
03

Calculate the Total Energy in the Solenoid's Magnetic Field

The total energy \( U \) contained in the solenoid is the product of the energy density \( u \) and the volume \( V \) of the solenoid. The volume is \( V = A \times L \), where \( A = 0.500 \times 10^{-4} \text{ m}^2 \) and \( L = 0.25 \text{ m} \):\[ V = 0.500 \times 10^{-4} \times 0.25 \]Then calculate \( U = u \times V \).
04

Calculate the Inductance of the Solenoid

The inductance \( L_s \) of a solenoid is calculated using:\[ L_s = \mu_0 \frac{N^2 A}{L} \]Substitute \( \mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} \), \( N = 400 \), \( A = 0.500 \times 10^{-4} \text{ m}^2 \), and \( L = 0.25 \text{ m} \):\[ L_s = 4\pi \times 10^{-7} \times \frac{400^2 \times 0.500 \times 10^{-4}}{0.25} \]Calculate to find the inductance \( L_s \).

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Key Concepts

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

Energy Density in Magnetic Field
When a solenoid is energized, it creates a magnetic field within and around it. This magnetic field stores energy. The term "energy density" refers to how much energy is stored in a unit volume of space. In the case of magnetic fields, this is given by the equation:\[ u = \frac{B^2}{2\mu_0} \]where \( u \) is the energy density, \( B \) is the magnetic field, and \( \mu_0 \) is the permeability of free space, typically valued at \( 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \).
  • Energy Concept: The energy density shows how efficiently energy is stored in a magnetic field.
  • Uniform Field: For a uniform field like in a solenoid, the energy density is constant throughout the area it permeates.
  • Application: It's crucial for understanding the power and efficiency of devices like electric motors and transformers.
By plugging in the calculated magnetic field strength from earlier, you can find how much energy is stored per cubic meter inside your solenoid.
Total Energy in Solenoid
To find the total energy stored in the magnetic field of a solenoid, you need to calculate the product of the energy density and the volume of the solenoid. The volume can be determined using:\[ V = A \times L \]where \( A \) is the cross-sectional area and \( L \) is the length of the solenoid. For our example, the area is 0.500 cm虏, which must be converted to meters squared (\(0.500 \times 10^{-4} \, \text{m}^2\)) to maintain consistency with SI units.
  • Volume Calculation: Calculate by multiplying the area by the length.
  • Total Energy: Use \( U = u \times V \) where \( u \) is the energy density, to find the total energy contained.
This total energy reflects how much energy the solenoid can store in its magnetic field, which is important for circuits involving inductive components, such as those in inductors and transformers.
Inductance of Solenoid
Inductance is a property of solenoids that describes their ability to resist changes in electric current flowing through them. This property is pivotal in creating electromagnets and comes in handy in various electrical applications. The formula for the inductance \( L_s \) of a solenoid is:\[ L_s = \mu_0 \frac{N^2 A}{L} \]where \( N \) is the number of wire turns, \( A \) is the cross-sectional area, and \( L \) is the length. \( \mu_0 \) again represents the permeability of free space.
  • Proportional Factors: Inductance increases with more turns and a larger area, while it decreases with a longer solenoid length.
  • Applications: Inductance is a key parameter in designing circuits that include coils and inductors, such as in radios and transformers.
Understanding inductance helps in the efficient design and application of various electrical devices that rely on controlled electromagnetic fields.

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Most popular questions from this chapter

A long, straight solenoid has 800 turns. When the current in the solenoid is 2.90 A, the average flux through each turn of the solenoid is \(3.25 \times 10^{-3}\) Wb. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal 6.20 mV?

A 15.0-\(\mu\)F capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280-mH inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time \(t = 0\) ms (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t = 1.30\) ms.

A 18.0-\(\mu\)F capacitor is placed across a 22.5-V battery for several seconds and is then connected across a 12.0-mH inductor that has no appreciable resistance. (a) After the capacitor and inductor are connected together, find the maximum current in the circuit. When the current is a maximum, what is the charge on the capacitor? (b) How long after the capacitor and inductor are connected together does it take for the capacitor to be completely discharged for the first time? For the second time? (c) Sketch graphs of the charge on the capacitor plates and the current through the inductor as functions of time.

An \(L\)-\(R\)-\(C\) series circuit has \(L = 0.600\) H and \(C = 3.00 \, \mu \mathrm{F}\). (a) Calculate the angular frequency of oscillation for the circuit when \(R =\) 0. (b) What value of \(R\) gives critical damping? (c) What is the oscillation frequency \(\omega'\) when \(R\) has half of the value that produces critical damping?

It has been proposed to use large inductors as energy storage devices. (a) How much electrical energy is converted to light and thermal energy by a 150-W light bulb in one day? (b) If the amount of energy calculated in part (a) is stored in an inductor in which the current is 80.0 A, what is the inductance?
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