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Chapter 30: Problem 6

A toroidal solenoid has 500 turns, cross-sectional area \(6.25 \mathrm{~cm}^{2}\) and mean radius \(4.00 \mathrm{~cm}\). (a) Calculate the coil's self-inductance. (b) If the current decreases uniformly from \(5.00 \mathrm{~A}\) to \(2.00 \mathrm{~A}\) in \(3.00 \mathrm{~ms},\) calculate the sclf- induced emf in the coil. (c) The current is dirccted from terminal \(a\) of the coil to terminal \(b .\) Is the direction of the induced emf from \(a\) to \(b\) or from \(b\) to \(a\) ? $

Short Answer

Expert verified
The calculations show that the self-inductance \(L\) of the coil is \(0.8 \times 10^{-3}\) H. The self-induced emf is \(800 V\) and it is from terminal \(b\) to terminal \(a\) due to Lenz's law.

Step by step solution

01

Calculate the Coil's Self-Inductance

The coil's self-inductance, also known as the magnetic inductance, can be calculated using the formula \[L= \mu_0 \frac{N^2A}{l}\] where \(L\) is the self-inductance, \(N\) is the number of turns, \(A\) is the cross-sectional area and \(l\) is the length of the coil, and \(\mu_0\) is the permeability of free space. Here, \(N=500\), \(A=6.25 \, cm^2 = 6.25 \times 10^{-4} \, m^2\) (converted from cm^2 to m^2 for SI units), and \(\mu_0 = 4 \pi \times 10^{-7} T m /A\). Given the solenoid is toroidal, the length \(l\) is the circumference of the solenoid, i.e, \(l = 2 \pi r = 2 \pi \times 0.04 \, m = 0.08 \pi \, m\).
02

Calculate the Self-Induced EMF

The self-induced emf in the coil can be calculated using Faraday's law of electromagnetic induction, which states that any change in the magnetic environment of a coil of wire will cause a voltage (emf) to be 'induced' in the coil. Mathematically, it is given by \[ emf = -L \frac{dI}{dt} \] where \(L\) is the self-inductance, \(dI\) is the change in current, and \(dt\) is the change in time. Here, \(-L\) signifies that the induced emf and the change in current have opposite signs. Given \(I\) decreases uniformly from 5A to 2A in 3ms, \(\frac{dI}{dt} = \frac{(2-5) A}{(3 \times 10^{-3}) s} = -1000 A/s\)
03

Determine the Direction of the Induced EMF

The direction of the self-induced emf can be determined through Lenz's law, which states that the direction of induced current is such that it opposes the change in current that produced it. As the given current decreases and is directed from terminal \(a\) to terminal \(b\), the direction of induced emf is from terminal \(b\) to terminal \(a\) (opposite to the direction of the decreasing current).
04

Computing the Values

Now, substituting all the given parameters into their respective formulas obtained from step 1 and step 2, calculate the coil's self-inductance \(L\) and the self-induced emf using the appropriate expressions.

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Key Concepts

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

Toroidal Solenoid
A toroidal solenoid is a fascinating component in electromagnetism. Imagine a doughnut-shaped coil of wire that consists of several tightly wound turns. This configuration is different from a regular solenoid, as there are no ends to the toroid. This design aids in containing the magnetic field inside the coil, reducing loss.

The key parts of a toroidal solenoid include:
  • Number of Turns ( N ): This is how many times the wire wraps around the core of the toroid. In this exercise, the toroid has 500 turns.
  • Mean Radius ( r ): The average radius of the circle formed by the center of the wire. The mean radius is 4.00 cm in our example.
  • Cross-Sectional Area ( A ): This is the area of the coil's cross-section through which the magnetic field passes. In our example, it is 6.25 cm虏, which needs to be converted into square meters for calculations.
Understanding these parts is crucial, as they help in calculating the coil's self-inductance, a measure of how effectively the coil induces emf in itself due to a changing current.
Faraday's Law of Electromagnetic Induction
Faraday's Law is a cornerstone principle of electromagnetism, revealing how electric currents can be generated from changing magnetic fields. According to this law, a voltage (emf) is induced in a circuit whenever there is a change in the magnetic environment of that circuit.

The law is simply expressed as:\[emf = -\frac{d\Phi_B}{dt}\]Where \(d\Phi_B\) is the change in magnetic flux over time. The negative sign demonstrates the direction of the induced emf, which follows Lenz's law to oppose the change in flux.
  • In Step 2 of our solution, we use this principle to determine the self-induced emf in the coil.
  • A decrease in the current results in a change in magnetic flux, thus creating an emf.
Faraday's Law connects the electric and magnetic worlds, showing how energy can be transformed from magnetic fields into electric voltages.
Lenz's Law
Lenz's Law provides clarity on the direction of the induced current or emf. It states that the induced emf will generate a current that opposes the change in magnetic flux that produced it. This law is a manifestation of the conservation of energy.

In our exercise, Lenz's Law helps determine that:
  • The current decreases from 5A to 2A, causing the magnetic field in the toroidal solenoid to decrease.
  • The direction of the induced emf must oppose this decrease.
  • Therefore, if the current is directed from terminal \(a\) to terminal \(b\), the induced emf will be from \(b\) to \(a\) to counter the reducing current.
This principle is crucial in applications ranging from electric generators to transformers, where it's essential to understand how circuits will respond to varying currents or magnetic fields.
Electromagnetic Induction
Electromagnetic induction refers to the process of generating an electric current from a changing magnetic field. This phenomenon underlies the operation of electrical devices like transformers and generators. The concept was primarily developed through the works of Michael Faraday and Joseph Henry.

Key elements of electromagnetic induction include:
  • Induced Emf: This is the voltage generated in a circuit due to changes in magnetic fields.
  • Magnetic Flux: Total magnetic field passing through a given area.
  • The Loop's Orientation: The angle of the loop relative to the magnetic field affects the amount of flux and thus the induced emf.
Understanding electromagnetic induction is vital, as it is the foundation for converting mechanical energy into electrical energy efficiently.

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Most popular questions from this chapter

A long. straight solenoid has 800 turns. When the current in the solenoid is \(2.90 \mathrm{~A}\), the average flux through each turn of the solenoid is \(3.25 \times 10^{-3} \mathrm{~Wb}\). What must be the magnitude of the rate of change of the current in order for the sclf-induced cmf to cqual \(6.20 \mathrm{mV} ?\)

It is possible to make your own inductor by winding wire around a cylinder, such as a pcncil. Assume you have a spool of AWG 20 copper wire, which has a diameter of \(0.812 \mathrm{~mm}\). (a) Estimate the diameter of a pencil. (b) Estimate how many times can you tightly wrap AWG 20 copper wire around a pencil to form a solenoid with a length of \(4.0 \mathrm{~cm}\). (c) Estimate the inductance of this solcnoid by assuming the magnetic field inside is constant. (d) If a current of 1.0 A flows through this solenoid, how much magnetic energy will be stored inside?

Two toroidal solenoids are wound around the same form so that the magnetic ficld of one passes through the turns of the other. Solenoid 1 has 700 turns, and solenoid 2 has 400 turns. When the current in solenoid 1 is \(6.52 \mathrm{~A},\) the average flux through each turn of solenoid 2 is \(0.0320 \mathrm{~Wb}\). (a) What is the mutual inductance of the pair of solcnoids? (b) When the current in solcnoid 2 is 2.54 A. what is the average flux through each turn of solenoid \(1 ?\)

{A} 6.40 \mathrm{nF} \text { capacitor is charged to } 24.0 \mathrm{~V} \text { and then discon- } nected from the battery in the circuit and connected in series with a coil that has \(L=0.0660 \mathrm{H}\) and negligible resistance. After the circuit has been completed, there are current oscillations. (a) At an instant when the charge of the capacitor is \(0.0800 \mu \mathrm{C}\), how much cnergy is stored in the capacitor and in the inductor, and what is the current in the inductor? (b) At the instant when the charge on the capacitor is \(0.0800 \mu \mathrm{C},\) what are the voltages across the capacitor and across the inductor, and what is the rate at which current in the inductor is changing?

CP CALC A Coaxial Cable. A small solid conductor with radius \(a\) is supported by insulating, nonmagnetic disks on the axis of a thin-walled tube with inner radius \(b\). The inner and outer conductors carry equal currents \(i\) in opposite directions. (a) Use Ampere's law to find the magnetic field at any point in the volume between the conductors. (b) Write the expression for the flux \(d \Phi_{B}\) through a narrow strip of length \(I\) parallel to the axis, of width \(d r\), at a distance \(r\) from the axis of the cable and lying in a planc containing the axis. (c) Intcgrate your expression from part (b) over the volume between the two conductors to find the total flux produced by a current \(i\) in the central conductor. (d) Show that the inductance of a length \(/\) of the cable is $$ I_{.}=l \frac{\mu_{0}}{2 \pi} \ln \left(\frac{b}{a}\right) $$ (e) Use Eq. (30.9) to calculate the energy stored in the magnetic field for a length \(l\) of the cable.
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