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Find the self-inductance per unit length of a long solenoid, of radius \(R\), carrying \(n\) turns per unit length.

Short Answer

Expert verified
The self-inductance per unit length is \( \mu_0 n^2 \pi R^2 \).

Step by step solution

01

Understanding the Solenoid

A solenoid is a long coil of wire wrapped in tight turns. It creates a magnetic field when an electric current passes through it. We are tasked with finding the self-inductance per unit length of this solenoid. To simplify, assume it's an ideal solenoid with uniform magnetic field inside.
02

Magnetic Field in Solenoid

The magnetic field inside an ideal solenoid is given by the formula \[ B = \mu_0 n I \]where \( B \) is the magnetic field, \( \mu_0 \) is the permeability of free space (\(4\pi \times 10^{-7} \ \text{Tm/A}\)), \( n \) is the number of turns per unit length, and \( I \) is the current through the solenoid.
03

Magnetic Flux Through the Solenoid

The magnetic flux through a single tightly wound loop in the solenoid is \[ \Phi = B \times \text{Area} = \mu_0 n I \pi R^2 \]where \( R \) is the radius of the solenoid.
04

Total Magnetic Flux Through the Solenoid

The total magnetic flux through \( n \) turns in a unit length of the solenoid is \[ \Phi_{total} = n \Phi = n (\mu_0 n I \pi R^2) = \mu_0 n^2 I \pi R^2 \]
05

Expression for Self-Inductance Per Unit Length

Self-inductance per unit length, \( L' \), is defined as the ratio of magnetic flux linkage to current. Hence, \[ L' = \frac{\Phi_{total}}{I} = \mu_0 n^2 \pi R^2 \] This is the self-inductance per unit length of the solenoid.

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Key Concepts

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

Solenoid
A solenoid is a fascinating component in electromagnetism. It's essentially a coil, typically made of many loops of wire, all aligned along a straight axis. When an electric current flows through this coil, it creates a magnetic field. This principle is due to the movement of electric charge, which naturally generates magnetism.

A great feature of a solenoid is how it can produce a nearly uniform magnetic field inside its core when tightly wound and long enough. This is why solenoids are often used in devices that need controlled magnetic environments or magnetic fields. It's crucial in applications from simple electromagnets to more complex ones like MRI machines and electric motors.
  • Long solenoids ensure a consistent inner field due to their geometry.
  • Turns per unit length ( ) directly influence the field strength.
  • The diameter or radius ( R) affects how the field spreads out.
Understanding solenoids helps in appreciating how electric circuits can control magnetic forces.
Magnetic Field
The magnetic field inside a solenoid is a vital concept. It's the region where magnetic forces act. For an ideal solenoid—a hypothetical model for simplification—this field is uniform. You can calculate it using the formula: \[ B = \mu_0 n I \]

This equation tells us essential aspects of the solenoid's magnetic nature:
  • \( B \) is the magnetic field strength.
  • \( \mu_0 \) is the permeability of free space, a constant measuring how easily a magnetic field can penetrate a material.
  • \( n \) represents the number of turns of the coil per unit length.
  • \( I \) is the electrical current flowing through the solenoid.
This magnetic field inside is crucial for the solenoid's practical uses, as it forms the basis for many electromechanical devices.
Magnetic Flux
Magnetic flux is an integral concept when dealing with solenoids. It can be understood as the measure of the magnetic field through a given area. Think of it as how much magnetic field "flow" passes through an area surface, described by the formula:\[ \Phi = B \times \text{Area} = \mu_0 n I \pi R^2 \]

Here, a few critical factors determine the flux:
  • The magnetic field strength (B).
  • The cross-sectional area of the solenoid (\( \pi R^2 \)).
Combining these gives the total magnetic flux through the solenoid, which in turn affects inductance. Changes in magnetic flux lead to induced electromotive forces (EMF), a principle central to transformer and generator technologies.
Permeability of Free Space
Permeability of free space, symbolized as \( \mu_0 \), plays a key role in determining how a magnetic field behaves in different environments. It's a constant value of approximately \( 4\pi \times 10^{-7} \, \text{Tm/A} \), and it determines how well the magnetic field lines can penetrate a vacuum.

In the context of a solenoid, \( \mu_0 \) helps define both the field strength and related magnetic phenomena, because it affects the propagation of magnetic fields. The constant shows up in key equations, such as the one that defines the magnetic field in a solenoid: \[ B = \mu_0 n I \] In understanding solenoids, it's crucial to use \( \mu_0 \) to accurately determine the effects of fields in vacuum environments. It lays the groundwork for interpreting electromagnetic interactions in and out of solenoids.

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Most popular questions from this chapter

A toroidal coil has a rectangular cross section, with inner radius \(a\), outer radius \(a+w\), and height \(h\). It carries a total of \(N\) tightly wound turns, and the current is increasing at a constant rate \((d l / d t=k)\). If \(w\) and \(h\) are both much less than \(a\), find the electric field at a point \(z\) above the center of the toroid.

Sea water at frequency \(v=4 \times 10^{8} \mathrm{~Hz}\) has permittivity \(\epsilon=81 \epsilon_{0}\), permeability \(\mu=\mu_{0}\), and resistivity \(\rho=0.23 \Omega \cdot \mathrm{m}\). What is the ratio of conduction current to displacement current?

A square loop, side \(a\), resistance \(R\), lies a distance \(s\) from an infinite straight wire that carries current \(I\) (Fig. 7.28). Now someone cuts the wire, so that \(I\) drops to zero. In what direction does the induced current in the square loop flow, and what total charge passes a given point in the loop during the time this current flows? If you doa't like the scissors model. turn the current down gradually: $$I(t)=\left\\{\begin{array}{ll}(1-\alpha t) l, & \text { for } 0 \leq t \leq 1 / \alpha \\ 0, & \text { for } t>1 / \alpha\end{array}\right.$$

A perfectly conducting spherical shell of radius \(a\) rotates about the \(z\) axis with angular velocity \(\omega\), in a uniform magnetic field \(\mathbf{B}=B_{0}\). \(\mathbf{z}\). Calculate the emf developed between the "north pole" and the equator, [Answer: \(\left.\frac{1}{2} B_{0} \omega a^{2}\right]\)

A square loop of wire (side \(a\) ) lies on a table, a distance \(s\) from a very long straight wire, which carries a current \(I\), as shown in Fig. \(7.17\). (a) Find the flux of \(\mathrm{B}\) through the loop. (b) If someone now pulls the loop directly away from the wire, at speed \(v\), what emf is generated? In what direction (clockwise or counterclockwise) does the current flow? (c) What if the loop is puiled to the right at speed \(v\), instead of away?

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