Problem 36 - College Physics

Question

Self-inductance of a solenoid. A long, straight solenoid has \(N\) turns, a uniform cross-sectional area \(A,\) and length \(l .\) Use the definition of self- inductance expressed by Equation 21.13 to show that the inductance of this solenoid is given approximately 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 of the solenoid. For this reason, your answer is actually an upper limit on the inductance.)

Step-by-Step Solution

Answer

The inductance of the solenoid is approximately \( L = \frac{\mu_0 N^2 A}{l} \).

Step 1: Recall the Definition of Self-Inductance

The self-inductance \( L \) of a solenoid can be defined as the ratio of the magnetic flux \( \Phi_B \) through the solenoid to the current \( I \) producing it. That is, \( L = \frac{N\Phi_B}{I} \), where \( \Phi_B \) is the magnetic flux through each turn and \( N \) is the number of turns.

Step 2: Express Magnetic Flux

Each turn of the solenoid encloses a magnetic field \( B \), which is uniform inside the solenoid. The magnetic flux \( \Phi_B \) through one turn is given by \( \Phi_B = B \cdot A \), where \( A \) is the cross-sectional area of the solenoid.

Step 3: Calculate Magnetic Field Inside the Solenoid

The magnetic field \( B \) inside a long solenoid is defined as \( B = \mu_0 \frac{NI}{l} \), where \( \mu_0 \) is the permeability of free space, \( N \) is the number of turns, \( I \) is the current, and \( l \) is the length of the solenoid.

Step 4: Substitute Magnetic Field into Flux Expression

Substitute \( B = \mu_0 \frac{NI}{l} \) into the expression for magnetic flux: \( \Phi_B = \left (\mu_0 \frac{NI}{l} \right) A = \mu_0 \frac{NAI}{l} \).

Step 5: Derive the Expression for Self-Inductance

Substituting \( \Phi_B = \mu_0 \frac{NAI}{l} \) into the definition of self-inductance, we get \( L = \frac{N (\mu_0 \frac{NAI}{l})}{I} = \frac{\mu_0 N^2 A I}{l \cdot I} = \frac{\mu_0 N^2 A}{l} \).

Key Concepts

Self-Inductance

Self-inductance is a fundamental concept in electromagnetism. It measures how a coil or solenoid resists changes in the electric current passing through it. Basically, when the current flowing through a coil changes, it alters the magnetic field around that coil. This change, in turn, induces an electromotive force (EMF) in the coil itself. The value of self-inductance, denoted by the symbol \( L \), shows how effectively a coil generates an induced EMF in response to changes in current.

Simplifying, self-inductance is the ratio of magnetic flux linked with the coil to the current creating that flux: \( L = \frac{N \Phi_B}{I} \).

In this formula:

• \( N \) is the number of turns or loops in the coil.
• \( \Phi_B \) represents the magnetic flux through each turn.
• \( I \) is the current.

The unit of self-inductance is Henry (H), which indicates the amount of inductance needed to induce one volt of EMF by a change of one ampere per second of current.

Magnetic Field

A magnetic field is a vector field around a magnet, electric current, or changing electric field, where magnetic forces are observable. Inside a solenoid, the magnetic field is particularly interesting because it is uniform and consistent, making calculations simpler and highly useful for applications like electromagnets.

The magnetic field inside a solenoid can be expressed using the formula \( B = \mu_0 \frac{NI}{l} \).

Where:

• \( B \) is the magnetic field strength.
• \( \mu_0 \) is the permeability of free space (a constant).
• \( N \) stands for the number of turns in the solenoid.
• \( I \) is the electric current flowing through it.
• \( l \) is the length of the solenoid.

Outside the solenoid, the magnetic field is negligible or nearly zero, which allows all field lines to focus effectively within it. This characteristic is widely used in creating strong magnetic fields in controlled environments.

Solenoid

A solenoid is a specific type of coil used extensively in electromagnetics and electrical engineering to create a controlled magnetic field. It consists of a long wire wound in many loops, forming a cylindrical shape. When electricity passes through each loop, it generates magnetic field lines parallel to the solenoid's axis, creating a concentrated and uniform magnetic field inside.

The solenoid's effectiveness at producing a uniform magnetic field makes it ideal for use in various applications such as:

• Electromagnets
• Magnetic resonance imaging (MRI)
• Actuators in engines
• Switches and relays

The properties of its magnetic field make it a practical tool for many technologies that need predictable and manageable magnetic fields. The design of a solenoid allows it to convert electrical energy into mechanical energy efficiently, which is why it is central to many mechanical and electronic systems.

Magnetic Flux

Magnetic flux quantifies the total magnetic field passing through a given area. It is essentially a measure of the number of magnetic field lines intersecting that surface. To put it simply, magnetic flux gives you an idea of how much 'magnetic stuff' is passing through a certain area.

In the context of a solenoid, we express magnetic flux as \( \Phi_B = B \cdot A \), where:

• \( \Phi_B \) is the magnetic flux.
• \( B \) is the magnetic field strength.
• \( A \) is the cross-sectional area through which the field lines pass.

This concept is central to the understanding of not just solenoids, but also broader electromagnetic applications, such as transformers and inductors. It helps in accounting for how changes in a magnetic field affect the creation of an electromotive force, central to Faraday's laws of electromagnetic induction.