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Inductance of a Solenoid. A long, straight solenoid has \(N\) turms, uniform cross-sectional area \(A,\) and length \(l .\) Show that the inductance of this solenoid is given by the equation \(L=\mu_{0} A N^{2} / L\) Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.)

Short Answer

Expert verified
The inductance of the solenoid is \(L = \mu_0 A N^2 / l\).

Step by step solution

01

Understand the Problem

We need to find the inductance \( L \) of a solenoid, which is characterized by its number of turns \( N \), cross-sectional area \( A \), and length \( l \). The inductance formula given is \( L = \mu_0 AN^2/l \). Here, \( \mu_0 \) is the magnetic constant or permeability of free space. The goal is to derive this formula based on the properties of the solenoid.
02

Identify the Magnetic Field

The magnetic field inside a long solenoid is given by the formula \( B = \mu_0 n I \), where \( n = \frac{N}{l} \) is the number of turns per unit length, and \( I \) is the current flowing through the solenoid. So, \( B = \mu_0 \frac{N}{l} I \).
03

Calculate the Magnetic Flux

The magnetic flux through one turn of the solenoid is \( \Phi = B \cdot A = (\mu_0 \frac{N}{l} I) \cdot A \). Since the solenoid has \( N \) turns, the total magnetic flux linkage is \( \lambda = N \Phi = \mu_0 \frac{N^2}{l} A I \).
04

Relate Flux Linkage to Inductance

The inductance \( L \) is defined as the ratio of the total magnetic flux linkage \( \lambda \) to the current \( I \). Thus, \( L = \frac{\lambda}{I} = \frac{\mu_0 N^2 A}{l} \). This matches the given formula for the inductance of the solenoid.

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Key Concepts

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

Magnetic Field
A magnetic field is an invisible force field that surrounds a magnetic object or electrical current. It is a vector field, which means it has both direction and magnitude. In the context of a solenoid, which is a coil of wire, the magnetic field inside is due to the flow of electric current through the wire coils. This field can be calculated using the formula for a solenoid:
  • The magnetic field inside a solenoid: \( B = \mu_0 \frac{N}{l} I \),where \( \mu_0 \) represents the permeability of free space, \( N \) is the number of turns, \( l \) is the length of the solenoid, and \( I \) is the current.
The solenoid's field is nearly uniform inside, with negligible field outside. This uniformity makes it easy to predict and use in practical applications, such as electromagnets and inductors.
Remember, the assumption of zero magnetic field outside is simplification for calculations and is termed as the ideal behavior of long solenoids.
Magnetic Flux
Magnetic flux quantifies the amount of magnetic field passing through a surface area. It tells us how much field lines penetrate a certain area, expressed with the formula:
  • Magnetic flux, \( \Phi = B \cdot A \), where \( B \) is the magnetic field and \( A \) is the area perpendicular to the field.
In a solenoid, we calculate the flux through one loop and then multiply by the total number of turns to get the total magnetic flux linkage (\( \lambda \)). The flux is linked to areas where magnetic force has an effect. For inductance, this is crucial as it determines how much field is created by electrical currents in coils.
Each turn contributes to the overall linkage, enhancing the total field through the solenoid's core.
Permeability of Free Space
Permeability of free space, often represented as \( \mu_0 \), is a constant that measures the ability of a vacuum to allow the formation of a magnetic field. In SI units, it is valued at approximately \( 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \) (Tesla meter per Ampere). This constant is crucial in calculating the inductance of a solenoid because it directly relates magnetic field strength to current and coil geometry. It's used in the inductance formula:
  • Inductance \( L = \mu_0 \frac{N^2 A}{l} \), emphasizing how magnetic flux density not only depends on the solenoid's material and construction but also on this universal constant.
\( \mu_0 \) helps determine how easily magnetic field lines can form between the coils when current flows, impacting the solenoid's effectiveness in many electromagnetic applications.

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Most popular questions from this chapter

Two coils have mutual inductance \(M=3.25 \times 10^{-4} \mathrm{H}\) . The current \(i_{1}\) in the first coil increases at a uniform rate of 830 \(\mathrm{A} / \mathrm{s}\) . (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is 4.18 \(\mathrm{pF}\) . (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L-C\) circuit is \(1600 \times 10^{3} \mathrm{Hz}\) , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3} \mathrm{Hz}\) . What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

An inductor used in a de power supply has an inductance of 12.0 \(\mathrm{H}\) and a resistance of \(180 \Omega .\) It carries a current of 0.300 \(\mathrm{A}\) . (a) What is the energy stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor?(c) Does your answer to part (b) mean that the magnetic-field energy is decreasing with time? Explain.

An Electromagnetic CarAlarm. Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3500 \(\mathrm{Hz}\) . To do this, the car-alarm circuitry must produce an altermating electric current of the same frequency. That's why your design includes an inductor and a capacitor in series. The maximum voltage across the capacitor is to be 12.0 \(\mathrm{V}\) (the same voltage as the car battery). To produce a sufficiently loud sound, the capacitor must store 0.0160 \(\mathrm{J}\) of energy. What values of capacitance and inductance should you choose for your car-alarm circuit?

A \(15.0-\Omega\) resistor and a coil are connected in series with a 6.30-V battery with negligible internal resistance and a closed switch. (a) At 200 ms after the switch is opened the current has decayed to 0.210 A. Calculate the inductance of the coil. (b) Calculate the time constant of the circuit. (c) How long after the switch is closed will the current reach 1.00\(\%\) of its original value?

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