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Chapter 22: Problem 56

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, so it has virtually the same radius as the solenoid. What is the mutual inductance of this system?

Short Answer

Expert verified
The mutual inductance of the system is approximately \(1.00 \times 10^{-4} \, ext{H (henries)}\).

Step by step solution

01

Understand the Concept of Mutual Inductance

Mutual inductance occurs when a change in current in one coil induces an electromotive force (EMF) in another coil that is close by. For two closely spaced coils, the mutual inductance, \(M\), is given by \(M = \frac{N_2 \Phi}{I_1}\) where \(N_2\) is the number of turns of the second coil, \(\Phi\) is the magnetic flux through one loop of the coil due to the current \(I_1\) in the solenoid.
02

Determine Magnetic Field Inside the Solenoid

The magnetic field inside a long solenoid is given by \(B = \mu_0 n I\), where \(\mu_0 = 4\pi \times 10^{-7} \ ext{T} ext{m/A}\) is the permeability of free space, \(n\) is the number of turns per meter, and \(I\) is the current. Here, \(n = 1750\) turns/m.
03

Calculate Magnetic Flux

The magnetic flux \(\Phi\) through the coil is given by \(\Phi = B \cdot A\), where \(A\) is the cross-sectional area of the solenoid, \(A = \pi r^2\) with \(r = 0.0180 \, ext{m}\). Thus, \(\Phi = (\mu_0 n I) \cdot (\pi r^2)\).
04

Substitute Flux into Mutual Inductance Formula

Use the given \(N_2 = 125\) to calculate the mutual inductance: \(M = \frac{N_2 \Phi}{I_1} = N_2 \cdot (\mu_0 n \cdot \pi r^2)\). Simplify to \(M = 125 \cdot (4\pi \times 10^{-7} \ ext{T} ext{m/A} \times 1750 \, ext{turns/m} \times \pi (0.0180 \, ext{m})^2)\).
05

Calculate the Mutual Inductance

Perform the calculations: \(M = 125 \times (4\pi \times 10^{-7} \times 1750 \times 3.1416 \times (0.0180)^2)\). Multiply these values to find \(M\).
06

Complete the Calculation

After performing the multiplication and simplification, the mutual inductance \(M\) is approximately \(1.00 \times 10^{-4} \, ext{H (henries)}\).

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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 long coil of wire, often wrapped tightly in the shape of a cylinder. When an electric current flows through the solenoid, it generates a magnetic field. This magnetic field is strongest inside the coil and is uniform in direction.
Solenoids are useful because they can create controlled magnetic fields and are often used in applications like electromagnets and inductors. The strength of the magnetic field inside a solenoid is determined by:
  • Number of turns per unit length (\( n \)).
  • The electric current (\( I \)) flowing through the wire.
The formula to calculate the magnetic field inside a solenoid is given by:\[B = \mu_0 n I\]where \( \mu_0 \)is the permeability of free space. This controlled environment makes solenoids ideal for experiments involving magnetic fields.
Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, considering the strength and extent of a magnetic field. It is represented by the symbol \( \Phi \)and is measured in webers (\( Wb \)).
In a solenoid, the magnetic flux through a loop or a coil is calculated by multiplying the magnetic field with the cross-sectional area:\[\Phi = B \cdot A\]where \( A = \pi r^2 \)is the area of the circle and \( r \)is the radius of the solenoid.
This concept is crucial when you want to understand how much magnetic field passes through each coil of wire. It's especially important in determining mutual inductance, as changes in magnetic flux induce an electromotive force in neighboring coils.
Current-Carrying Coil
A current-carrying coil, like the one wrapped around the solenoid in our scenario, generates its own magnetic field. When wrapped around another coil, it can exhibit interesting electromagnetic properties such as mutual inductance.
This mutual influence happens when the current flowing through the solenoid induces a current or electromotive force (EMF) in the nearby coil due to changing magnetic fields. The number of turns in the coil and its proximity to the solenoid are crucial factors determining the effect.
The interplay between the solenoid and coil is key in many electromagnetic devices, such as transformers and inductors, where energy is transferred efficiently between circuits.
Electromotive Force (EMF)
Electromotive force (EMF) is the energy provided per coulomb of charge by a power source or by a changing magnetic field. Even though it is called 'force,' it is actually a potential difference and is measured in volts.
In the context of mutual inductance, the changing current in the solenoid induces an EMF in the coil wrapped around it. The formula relating to this induced EMF is given by Faraday's Law of Induction:\[\text{EMF} = -N \frac{d\Phi}{dt}\]where \( N \)is the number of turns in the coil, and \( \frac{d\Phi}{dt} \)represents the rate of change of magnetic flux.
This induced EMF can lead to a current in the second coil, allowing energy transfer without any physical connection, which is a fundamental concept in transformer operation.

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Most popular questions from this chapter

In 1996, NASA performed an experiment called the Tethered Satellite experiment. In this experiment a \(2.0 \times 10^{4}-\mathrm{m}\) length of wire was let out by the space shuttle Atlantis to generate a motional emf. The shuttle had an orbital speed of \(7.6 \times 10^{3} \mathrm{m} / \mathrm{s},\) and the magnitude of the earth's magnetic field at the location of the wire was \(5.1 \times 10^{-5}\) T. If the wire had moved perpendicular to the earth's magnetic field, what would have been the motional emf generated between the ends of the wire?

A square loop of wire consisting of a single turn is perpendicular to a uniform magnetic field. The square loop is then re-formed into a circular loop, which also consists of a single turn and is also perpendicular to the same magnetic field. The magnetic flux that passes through the square loop is \(7.0 \times 10^{-3} \mathrm{Wb}\). What is the flux that passes through the circular loop?

A generator uses a coil that has 100 turns and a \(0.50-\mathrm{T}\) magnetic field. The frequency of this generator is \(60.0 \mathrm{Hz},\) and its emf has an rms value of \(120 \mathrm{V}\). Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

A flat coil of wire has an area \(A, N\) turns, and a resistance \(R .\) It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of \(90^{\circ},\) so that the normal becomes perpendicular to the magnetic field. The coil has an area of \(1.5 \times 10^{-3} \mathrm{m}^{2}, 50\) turns, and a resistance of \(140 \Omega .\) During the time while it is rotating, a charge of \(8.5 \times 10^{-5} \mathrm{C}\) flows in the coil. What is the magnitude of the magnetic field?

A circular coil of radius 0.11 m contains a single timm and is located in a constant magnetic field of magnitude \(0.27 \mathrm{T} .\) The magnetic field has the same direction as the normal to the plane of the coil. The radius increases to \(0.30 \mathrm{m}\) in a time of \(0.080 \mathrm{s} .\) Concepts: (i) Why is there an emf induced in the coil? (ii) Does the magnitude of the induced emf depend on whether the area is increasing or decreasing? Explain. (iii) What determines the amount of current induced in the coil? (iv) If the coil is cut so it is no longer one continuous piece, are there an induced emf and an induced current? Explain. Calculations: (a) Determine the magnitude of the emf induced in the coil. (b) The coil has a resistance of \(0.70 \Omega\). Find the magnitude of the induced current.
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