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Chapter 22: Problem 51

A long, current-carrying solenoid with an air core has 1750 turns per meter of length and a radius of \(0.0180 \mathrm{~m}\). A coil of 125 turns is wrapped tightly around the outside of the solenoid. What is the mutual inductance of this system?

Short Answer

Expert verified
The mutual inductance of the system is approximately \(2.67 \times 10^{-5} \text{ H}\).

Step by step solution

01

Determine Magnetic Field Inside Solenoid

For a solenoid, the magnetic field inside the solenoid is uniform and is given by the expression: \( B = \mu_0 n I \), where \( \mu_0 \) is the permeability of free space \( (4\pi \times 10^{-7} \text{ Tm/A}) \), \( n \) is the number of turns per meter in the solenoid, and \( I \) is the current. Since we are dealing with mutual inductance and the field is not dependent on \( I \) for the formula, we express the general form here. Thus, the magnetic field \( B = \mu_0 n \).
02

Calculate the Flux Through a Single Turn of the Coil

The magnetic flux \( \Phi \) through a single loop of the coil is given by \( \Phi = B \cdot A \), where \( A \) is the area of one loop of the solenoid. Since \( A = \pi r^2 \) (with \( r \) being the radius of the solenoid), we get \( \Phi = \mu_0 n \times \pi r^2 \).
03

Calculate the Total Flux Through the Coil

The total magnetic flux linked with the coil, which has \( N_2 \) turns (125 turns), is given by: \( \Phi_{ ext{total}} = N_2 \cdot \Phi \). Hence, \( \Phi_{ ext{total}} = 125 \cdot \mu_0 n \cdot \pi r^2 \).
04

Apply the Formula for Mutual Inductance

Mutual inductance \( M \) is defined by the relationship \( M = \frac{\Phi_{ ext{total}}}{I} \). Expressing this in terms of the previous steps gives \( M = 125 \cdot \mu_0 n \cdot \pi r^2 \). Substitute the known values: \( n = 1750 \), \( r = 0.0180 \text{ m} \), and \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \).
05

Calculate the Value of the Mutual Inductance

Substitute all the numbers into the expression for \( M \): \[ M = 125 \cdot (4\pi \times 10^{-7}) \cdot 1750 \cdot \pi (0.0180)^2 \]. Evaluate this expression to find \( M = 2.67 \times 10^{-5} \text{ H} \).

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Key Concepts

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

Solenoid
A solenoid is essentially a long coil of wire that is tightly wound in a helical pattern. It serves as a fundamental component in electromagnetism and is widely used in various applications such as inductors, electromagnets, and antennas. When an electric current passes through the solenoid, it generates a magnetic field inside it.
  • The solenoid's length and the number of turns per unit length (turn density) are crucial factors.
  • A tightly wound solenoid can create a uniform magnetic field within its interior.
In our problem, the solenoid has an air core, with 1750 turns per meter and a radius of 0.018 meters. This setup creates a stable and predictable magnetic environment, essential for calculating mutual inductance with a surrounding coil.
Magnetic Field
The magnetic field inside a solenoid is an essential concept when discussing mutual inductance. For a solenoid, the magnetic field is uniform and is calculated using the formula: \[ B = \mu_0 n I \] Here:
  • \(B\) represents the magnetic field strength.
  • \(\mu_0\) is the permeability of free space, a constant \( (4\pi \times 10^{-7} \text{ Tm/A}) \).
  • \(n\) is the number of turns per meter.
  • \(I\) is the electric current flowing through the solenoid.
In our problem, the expression is simplified to \( B = \mu_0 n \) since we are focusing on mutual inductance, which relates to changes in magnetic flux rather than the current directly. The uniform magnetic field inside the solenoid is what links the coil wrapped around it, enabling mutual inductance.
Coil
The coil in this exercise is wrapped tightly around the solenoid, which means it experiences the full effect of the solenoid's magnetic field. Wraps around the solenoid like this are often used to harness the magnetic field generated by the solenoid for purposes such as energy transfer.
  • The coil consists of a specific number of turns, which in this case is 125 turns.
  • The proximity and tight wrapping enhance the interaction between the coil and the solenoid.
This leads to an effective linkage in terms of magnetic flux. The coil essentially picks up the changes in the magnetic field, which induce an electromotive force (emf) through the process of electromagnetic induction. Therefore, the design of the coil around the solenoid is central to its operation in mutual inductance calculations.
Magnetic Flux
Magnetic flux is a key concept in understanding mutual inductance as it represents how much magnetic field passes through a given area.
  • It's denoted by \( \Phi \) and calculated using: \( \Phi = B \cdot A \).
  • \(B\) is the magnetic field and \(A\) is the area through which the magnetic field lines pass.
In the context of the solenoid, the area, \( A \), for a single loop is determined using \( \pi r^2 \), where \( r \) is the radius of the solenoid.
The total magnetic flux through the coil is a sum of the fluxes through all its turns. For 125 turns, it's \( \Phi_{\text{total}} = 125 \cdot \Phi \). This total flux is pivotal in calculating mutual inductance, which describes how effectively a change in current in the solenoid induces voltage across the coil. As such, understanding magnetic flux is critical for grasping the functioning of mutual inductance.

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Most popular questions from this chapter

Multiple-Concept Example 13 reviews the concepts used in this problem. A long solenoid (cross-sectional area \(=1.0 \times 10^{-6} \mathrm{~m}^{2}\), number of turns per unit length \(=2400\) turns \(/ \mathrm{m}\) ) is bent into a circular shape so it looks like a doughnut. This wire-wound doughnut is called a toroid. Assume that the diameter of the solenoid is small compared to the radius of the toroid, which is \(0.050 \mathrm{~m}\). Find the emf induced in the toroid when the current decreases to \(1.1\) A from \(2.5 \mathrm{~A}\) in a time of \(0.15 \mathrm{~s}\).

A \(120.0\) - \(\mathrm{V}\) motor draws a current of \(7.00 \mathrm{~A}\) when running at normal speed. The resistance of the armature wire is \(0.720 \Omega\). (a) Determine the back emf generated by the motor. (b) What is the current at the instant when the motor is just turned on and has not begun to rotate? (c) What series resistance must be added to limit the starting current to \(15.0 \mathrm{~A} ?\)

Interactive LearningWare 22.2 at reviews the fundamental approach in problems such as this. A constant magnetic field passes through a single rectangular loop whose dimensions are \(0.35 \mathrm{~m} \times 0.55 \mathrm{~m}\). The magnetic field has a magnitude of \(2.1 \mathrm{~T}\) and is inclined at an angle of \(65^{\circ}\) with respect to the normal to the plane of the loop. (a) If the magnetic field decreases to zero in a time of \(0.45 \mathrm{~s}\), what is the magnitude of the average emf induced in the loop? (b) If the magnetic field remains constant at its initial value of \(2.1 \mathrm{~T},\) what is the magnitude of the rate \(\Delta \mathrm{A} / \Delta t\) at which the area should change so that the average emf has the same magnitude as in part (a)?

Interactive Solution \(\underline{22.55}\) at offers one approach to problems such as this one. The secondary coil of a step-up transformer provides the voltage that operates an electrostatic air filter. The turns ratio of the transformer is 50: 1 . The primary coil is plugged into a standard \(120-\mathrm{V}\) outlet. The current in the secondary coil is \(1.7 \times 10^{-3} \mathrm{~A} .\) Find the power consumed by the air filter.

Magnetic resonance imaging (MRI) is a medical technique for producing pictures of the interior of the body. The patient is placed within a strong magnetic field. One safety concern is what would happen to the positively and negatively charged particles in the body fluids if an equipment failure caused the magnetic field to be shut off suddenly. An induced emf could cause these particles to flow, producing an electric current within the body. Suppose the largest surface of the body through which flux passes has an area of \(0.032 \mathrm{~m}^{2}\) and a normal that is parallel to a magnetic field of \(1.5 \mathrm{~T}\). Determine the smallest time period during which the field can be allowed to vanish if the magnitude of the average induced emf is to be kept less than \(0.010 \mathrm{~V}\).
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