Problem 15 - University Physics with Modern Physics

Question

Inductance of a Solenoid. (a) A long, straight sole- noid has \(N\) turns, uniform crossectional 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.) (b) A metallic laboratory spring is typically 5.00 \(\mathrm{cm}\) long and 0.150 \(\mathrm{cm}\) in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

Step-by-Step Solution

Answer

The solenoid's inductance formula is \(L = \mu_{0} A N^2 / l\). For the spring, \(L \approx 2.10 \times 10^{-6} \, \text{H}.\)

Step 1: Understand the parameters

Identify the given parameters for the solenoid and the spring. The solenoid has \(N\) turns, a cross-sectional area \(A\), and length \(l\). For the spring, it is 5.00 cm long with 0.150 cm diameter and 50 coils; we model this spring as a solenoid.

Step 2: Formulate the inductance formula

The inductance \(L\) of a solenoid is given by the formula \(L = \mu_{0} \cdot \frac{N^2 \cdot A}{l}\). Here, \(\mu_{0}\) is the permeability of free space, a constant \(\approx 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A}\).

Step 3: Derive for the Solenoid Using its Properties

Use Ampere's circuital law and the characteristics of a solenoid to derive the expression for its magnetic field \(B = \mu_{0} \cdot \frac{N}{l} \cdot I\). The energy stored in the solenoid relates directly to inductance: \(E = \frac{1}{2} L I^2 = \frac{B^2}{2\mu_{0}} \cdot A \cdot l\). Equating these and simplifying gives \(L = \mu_{0} \cdot \frac{N^2 \cdot A}{l}\).

Step 4: Dimensions and Cross-sectional Area of the Spring

For the spring, calculate the cross-sectional area \(A\) using \(A = \pi \cdot r^2\), where \(r = 0.150 \mathrm{cm} \div 2 = 0.075 \mathrm{cm}\). Convert \(r\) into meters for accuracy: \(r = 0.00075 \mathrm{m}\). Thus, \(A = \pi \cdot (0.00075)^2\).

Step 5: Calculate the Inductance of the Spring

Now use the solenoid inductance formula for the spring: \(L = \mu_{0} \cdot \frac{N^2 \cdot A}{l}\). Plug in \(N = 50\), \(A\) as calculated, and \(l = 0.05 \, \text{m}\). Maintain units in SI to ensure consistency.

Key Concepts

Magnetic Field

A magnetic field is a region around a magnet where magnetic forces are exerted. In the case of a solenoid, the magnetic field inside is strong, uniform, and directed along the axis of the solenoid. Outside the solenoid, the field is much weaker. This is because all the field lines inside are packed closely together, creating a dense and powerful field.

The strength of this magnetic field inside a solenoid can be calculated using the formula:
\[ B = \mu_{0} \cdot \frac{N}{l} \cdot I \]
where \( B \) is the magnetic field, \( \mu_{0} \) is the permeability of free space, \( N \) is the number of turns, \( l \) is the length of the solenoid, and \( I \) is the current passing through it.

This equation underlines the principle that more turns and higher current increase the solenoid's magnetic field.

Ampère's Circuital Law

Ampère's circuital law is a fundamental principle that relates magnetic fields to the currents that produce them. Specifically, it states that the line integral of the magnetic field around a closed loop is equal to the permeability of free space times the current passing through that loop. The mathematical expression for this is given by:
\[ \int B \cdot dl = \mu_{0} \cdot I_{enclosed} \]

Ampère’s law helps us understand how the magnetic field inside a solenoid can be uniform. Since a solenoid wraps current in multiple loops around a core, it creates a continuous loop of lines of the magnetic field, enhancing the field strength inside. By calculating this using Ampère's circuital law, we assume the simple case of the field being uniform throughout the solenoid, although it’s slightly less at the ends.

Using this law, the formula for magnetic field inside a solenoid can be derived which further assists in determining other quantities like inductance.

Self-Inductance

Self-inductance is a property of a solenoid or any coil that quantifies its ability to induce a voltage within itself when the current flowing through it changes. This induced voltage opposes changes in the current, according to Lenz's law.

In the context of a solenoid, its self-inductance \( L \) is given by the formula:
\[ L = \mu_{0} \cdot \frac{N^2 \cdot A}{l} \]

Here, \( \mu_{0} \) denotes the permeability of free space, \( N \) is the number of turns, \( A \) is the cross-sectional area, and \( l \) is the length. The higher the number of turns or the larger the area, the greater the self-inductance.

Self-inductance is essential in electrical circuits because it defines the ability of a device to store energy in its magnetic field and influences how circuits respond to changes in current.

Permeability of Free Space

Permeability of free space, denoted \( \mu_{0} \), is a physical constant that describes how a magnetic field is established in a vacuum. Its approximate value is \( \mu_{0} = 4\pi \times 10^{-7} \, \text{T}\cdot\text{m/A} \).

This constant appears in equations, such as the formula for the magnetic field inside a solenoid and in the calculation for self-inductance. It fundamentally represents the ability of free space to allow magnetic field lines to pass through it efficiently.

The importance of \( \mu_{0} \) in magnetic field calculations lies in its role as a scaling factor that adjusts the relationship between current and induced magnetic fields in the absence of other materials. This helps in predicting the behavior of solenoids and other electromagnetic devices accurately.