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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 self- induced emf in the coil. (c) The current is directed 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
(a) The self-inductance is \(0.49\,\text{mH}\). (b) The induced emf is \(0.49\,\text{V}\). (c) The induced emf direction is from \(a\) to \(b\).

Step by step solution

01

Understand the Problem

We need to find the self-inductance of a toroidal solenoid, and calculate the induced emf when the current changes uniformly over time. The direction of the induced emf should also be determined. Key given parameters: number of turns \(N = 500\), cross-sectional area \(A = 6.25\,\text{cm}^2\), mean radius \(r = 4.00\,\text{cm}\).
02

Convert Units

First, convert all given units to SI units. The cross-sectional area \(A\) is \(6.25\,\text{cm}^2 = 6.25 \times 10^{-4}\,\text{m}^2\). The mean radius \(r\) is \(4.00\,\text{cm} = 0.0400\,\text{m}\).
03

Calculate Self-Inductance

The formula for the self-inductance \(L\) of a toroidal solenoid is \(L = \mu_0\frac{N^2A}{2\pi r}\), where \(\mu_0\) is the permeability of free space, \(\mu_0 = 4\pi \times 10^{-7}\,\text{T m/A}\). Substitute the values: \[L = (4\pi \times 10^{-7}) \frac{(500)^2 \times 6.25 \times 10^{-4}}{2\pi \times 0.0400}\] Solve to find \(L = 4.9 \times 10^{-4}\,\text{H}\).
04

Calculate the Rate of Change of Current

The current changes from \(5.00\,\text{A}\) to \(2.00\,\text{A}\) in \(3.00\,\text{ms} = 3.00 \times 10^{-3} \text{s}\). Calculate the rate of change of current: \( \frac{dI}{dt} = \frac{2.00 - 5.00}{3.00 \times 10^{-3}} = -1000\,\text{A/s}\).
05

Calculate Self-Induced EMF

The self-induced emf \(\mathcal{E}\) is given by the formula \(\mathcal{E} = -L \frac{dI}{dt}\). Substitute \(L = 4.9 \times 10^{-4}\,\text{H}\) and \(\frac{dI}{dt} = -1000\,\text{A/s}\): \[\mathcal{E} = - (4.9 \times 10^{-4}) (-1000) = 0.49\,\text{V}\].
06

Determine the Direction of Induced EMF

The induced emf opposes the change in current. Since the current is decreasing from \(a\) to \(b\), the induced emf will try to increase the current in the same direction. Thus, the direction of the induced emf is from \(a\) to \(b\).

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Key Concepts

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

Self-inductance
Self-inductance is a fundamental concept in electromagnetism. It's the property of a coil or solenoid that allows it to resist changes in the electric current flowing through it. When the current in a coil changes, it induces a voltage (electromagnetic force or emf) in the same coil, which opposes the change in current. This phenomenon is described by Lenz's Law and quantified by the inductance, measured in henries (H).
Inductance depends on factors like the number of turns of the coil, the cross-sectional area of the coil, and the material's permeability that surrounds it. For a toroidal solenoid, which is a coil wound into a donut shape, the formula we use is:\[L = \mu_0 \frac{N^2 A}{2 \pi r}\]where \(L\) is the self-inductance, \(\mu_0\) is the permeability of free space, \(N\) is the number of turns, \(A\) is the cross-sectional area, and \(r\) is the mean radius.
This equation shows how the self-inductance increases with more turns, a larger area, and a material with higher permeability.
Toroidal solenoid
A toroidal solenoid is a special type of solenoid coil shaped like a torus, which is effectively a solenoid bent into a circle. This toroidal design is quite effective in electromagnetic applications because it contains nearly all of the magnetic field within its structure, reducing electromagnetic interference (EMI) with external devices.
Key features of a toroidal solenoid include:
  • High efficiency at containing magnetic fields.
  • Reduced electromagnetic interference compared to straight solenoids.
  • A consistent magnetic field, given its circular and closed structure.
In applications requiring minimal interference, such as sensitive electronic circuits, toroidal solenoids are highly valued. They are used in transformers, inductors, and magnetic sensors, among various other electronic components.
Induced emf
Induced emf (or electromotive force) arises when the magnetic field within a closed loop changes over time. In the context of a toroidal solenoid, when the current through the coil changes, it induces an emf due to the self-inductance of the coil. According to Faraday’s Law of electromagnetic induction, the induced emf in any closed circuit is equal to the negative rate of change of the magnetic flux through the circuit.
Induced emf is calculated using the formula:\[\mathcal{E} = -L \frac{dI}{dt}\]where \(\mathcal{E}\) is the induced emf, \(L\) is the self-inductance, and \(\frac{dI}{dt}\) is the rate of change of current. The negative sign is a representation of Lenz's Law, indicating that the induced emf always works to oppose the change in current that produced it. This is nature's way of conserving energy.
Rate of change of current
The rate of change of current refers to how quickly the current through a circuit varies with time. It is an essential factor in determining the induced emf. When there is a rapid change in current, it generally results in a higher induced emf.
To find the rate of change of current, you can use the formula:\[\frac{dI}{dt} = \frac{I_{final} - I_{initial}}{\Delta t}\]The formula above calculates how much the current has changed divided by the interval of time over which the change occurred. This concept is crucial in circuits involving inductors, like our toroidal solenoid example.
A noticeable familiar situation is the decrease of current, which in turn leads to a negative rate of change of current. This negative sign means the current is decreasing, which impacts the direction of the induced emf as per Lenz’s law, ultimately trying to sustain the original current flow.

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Most popular questions from this chapter

In an \(L-C\) circuit, \(L=85.0 \mathrm{mH}\) and \(C=3.20 \mu \mathrm{F} .\) During the oscillations the maximum current in the inductor is 0.850 \(\mathrm{mA}\) . (a) What is the maximum charge on the capacitor? (b) What is the magnitude of the charge on the capacitor at an instant when the current in the inductor has magnitude 0.500 \(\mathrm{mA}\) ?

An \(L-C\) circuit consists of a \(60.0-\mathrm{mH}\) inductor and a \(250-\mu \mathrm{F}\) capacitor. The initial charge on the capacitor is 6.00\(\mu \mathrm{C}\) , and the initial current in the inductor is zero. (a) What is the maximum voltage across the capacitor? (b) What is the maximum current in the inductor? (c) What is the maximum energy stored in the inductor? (d) When the current in the inductor has half its maximum value, what is the charge on the capacitor and what is the energy stored in the inductor?

An air-filled toroidal solenoid has a mean radius of 15.0 \(\mathrm{cm}\) and a cross-sectional area of 5.00 \(\mathrm{cm}^{2} .\) When the current is 12.0 \(\mathrm{A}\) , the energy stored is 0.390 \(\mathrm{J} .\) How many turns does the winding have?

A Radio Tuning Circuit. The minimum capacitance of a variable capacitor in a radio is 4.18 \(\mathrm{pF}\) . (a) What is the inductance of a coil connected to this capacitor if the oscillation frequency of the \(L\) circuit is \(1600 \times 10^{3} \mathrm{Hz}\) , corresponding to one end of the AM radio broadcast band, when the capacitor is set to its minimum capacitance? (b) The frequency at the other end of the broadcast band is \(540 \times 10^{3}\) Hz. What is the maximum capacitance of the capacitor if the oscillation frequency is adjustable over the range of the broadcast band?

A \(20.0-\mu \mathrm{F}\) capacitor is charged by a \(150.0-\mathrm{V}\) power supply, then disconnected from the power and connected in series with a \(0.280-\mathrm{mH}\) inductor. Calculate: (a) the oscillation frequency of the circuit; (b) the energy stored in the capacitor at time \(t=0 \mathrm{ms}\) (the moment of connection with the inductor); (c) the energy stored in the inductor at \(t=1.30 \mathrm{ms}\) .

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