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A coiled telephone cord forms a spiral with 70 turns, a diameter of \(1.30 \mathrm{cm},\) and an unstretched length of \(60.0 \mathrm{cm} .\) Determine the self-inductance of one conductor in the unstretched cord.

Short Answer

Expert verified
The self-inductance of the cord is calculated by the aforementioned steps, using the formula for the inductance of a solenoid and utilizing parameters such as the number of turns, length of the coil, and the diameter. Detailed calculations will provide the exact value in Henry.

Step by step solution

01

Calculating for number of turns per unit length (n)

The number of turns per unit length, \( n \), can be calculated as total number of turns divided by length of the cord. Hence \( n = \frac{Number of turns}{Length} = \frac{70}{60.0 \, cm} \). Since we want \( n \) in turns/meter, convert 60.0 cm to meters to get \( n = \frac{70}{0.60 \, m} \) turns/meter.
02

Calculating for Radius (r)

The radius (\( r \)) of the solenoid can be calculated from the given diameter. Thus, \( r = \frac{Diameter}{2} = \frac{1.30 \, cm}{2} \). Again convert radius to meter so get \( r = \frac{0.013 \, m}{2} \) meters.
03

Calculating Self-Inductance (L)

Now using the formula for the self-inductance of a solenoid, \( L = \mu_{0} \cdot n^2 \cdot \pi \cdot r^2 \cdot l \), where \( \mu_{0} \) is the permeability of free space \( (4\pi \times 10^{-7} \, T \cdot m/A) \). Substituting the values gives \( L = (4\pi \times 10^{-7} \, T \cdot m/A) \cdot (\frac{70}{0.60})^2 \cdot \pi \cdot (0.013/2) \cdot 0.60 \).
04

Simplifying the Expression for L

The expression for L, when simplified using a calculator will give the self-inductance of the coil in Henry (H).

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Key Concepts

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

Solenoid Inductance Calculation
Understanding the self-inductance of a solenoid is essential for various applications in electromagnetism. Self-inductance, often denoted as \'L\', quantifies the ability of a conductor to induce a voltage on itself when the current flowing through it changes. A solenoid, which is essentially a coil of wire, has significant self-inductance due to its shape and the number of turns.

When it comes to calculating the self-inductance of a solenoid, a straightforward formula is used: \[ L = \mu_0 \cdot n^2 \cdot \pi \cdot r^2 \cdot l \. \] Here, \( L \) represents the self-inductance measured in Henrys (H), \( \mu_0 \) is the magnetic permeability of free space, \( n \) is the number of turns per unit length, \( r \) is the radius of the solenoid, and \( l \) is the length of the solenoid.

For a coiled telephone cord with 70 turns, a 0.013 m diameter, and a 0.60 m length, the number of turns per unit length \((n)\) and the radius \((r)\) are computed first, followed by substituting these values into the inductance formula. It's crucial to ensure all units are consistent, typically using the SI system, for an accurate result.

Ultimately, precise computation is essential for applications where the inductance value is critical for performance, such as in tuning circuits or inductor design for electronic devices.
Magnetic Permeability
Magnetic permeability \((\mu)\) is a fundamental property that describes how a material responds to a magnetic field. Specifically, it indicates the ease with which a magnetic field can penetrate a medium. The permeability of free space, \(\mu_0\), is a constant value used in many electromagnetic calculations, including the inductance of a solenoid.

\(\mu_0\) has a standardized value of \(4\pi \times 10^{-7} \, T\cdot m/A\) (Teslas times meters per Ampere) and forms the basis of magnetic measurements in a vacuum. When dealing with solenoids, we typically assume they are in air, which has a permeability very close to that of a vacuum. Hence, \(\mu_0\) accurately reflects the medium\'s characteristics for calculating solenoid inductance.

In our example with the telephone cord, the permeability of free space is used for the self-inductance calculation. If the solenoid was in a different medium, like iron, the relative permeability of that medium would also need to be accounted for, significantly impacting the inductance.
Turns per Unit Length
The term \(\text{\'turns per unit length\'}\) is a crucial aspect when calculating the inductance of a solenoid. It represents the number of turns of the wire wound along the length of the solenoid. Mathematically, it is expressed as \( n \) and calculated using the following formula: \[ n = \frac{\text{Number of turns}}{\text{Length of solenoid}} \. \]

In the case of the telephone cord used as a solenoid with 70 turns and a length of 0.60 meters, turns per unit length is found by dividing the total number of turns by the solenoid's length, resulting in an \( n \) value expressed in units of turns per meter (turns/m).

The density of the coil\'s winding, or \( n \), directly affects the solenoid\'s inductance: the more turns per meter, the higher the inductance. It is vital that calculations are double-checked for accuracy, especially the conversion of units, as this directly influences the outcome of the inductance calculation. Understanding how this property influences magnetic fields and inductance helps in applications like electric motors, transformers, and antenna design where specific inductance values are essential.

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Most popular questions from this chapter

Two coils, held in fixed positions, have a mutual inductance of \(100 \mu \mathrm{H}\). What is the peak voltage in one when a sinusoidal current given by \(I(t)=(10.0 \mathrm{A}) \sin (1000 t)\) is in the other coil?

The energy of an \(R L C\) circuit decreases by \(1.00 \%\) during each oscillation when \(R=2.00 \Omega .\) If this resistance is removed, the resulting \(L C\) circuit oscillates at a frequency of \(1.00 \mathrm{kHz}\). Find the values of the inductance and the capacitance.

This problem extends the reasoning of Section \(26.4,\) Problem \(26.37,\) Example \(30.6,\) and Section 32.3. (a) Consider a capacitor with vacuum between its large, closely spaced, oppositely charged parallel plates. Show that the force on one plate can be accounted for by thinking of the electric field between the plates as exerting a "negative pressure" equal to the energy density of the electric field. (b) Consider two infinite plane sheets carrying electric currents in opposite directions with equal linear current densities \(J_{s} .\) Calculate the force per area acting on one sheet due to the magnetic field created by the other sheet. (c) Calculate the net magnetic field between the sheets and the field outside of the volume between them. (d) Calculate the energy density in the magnetic field between the sheets. (e) Show that the force on one sheet can be accounted for by thinking of the magnetic field between the sheets as exerting a positive pressure equal to its energy density. This result for magnetic pressure applies to all current configurations, not just to sheets of current.

A capacitor in a series \(L C\) circuit has an initial charge \(Q\) and is being discharged. Find, in terms of \(L\) and \(C\), the flux through each of the \(N\) turns in the coil, when the charge on the capacitor is \(Q / 2.\)

An air-core solenoid \(0.500 \mathrm{m}\) in length contains 1000 turns and has a cross-sectional area of \(1.00 \mathrm{cm}^{2} .\) (a) Ignoring end effects, find the self-inductance. (b) A secondary winding wrapped around the center of the solenoid has 100 turns. What is the mutual inductance? (c) The secondary winding carries a constant current of \(1.00 \mathrm{A},\) and the solenoid is connected to a load of \(1.00 \mathrm{k} \Omega .\) The constant current is suddenly stopped. How much charge flows through the load resistor?

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