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A toroidal solenoid has 500 turns, cross-sectional area 6.25 cm\(^2\), and mean radius 4.00 cm. (a) Calculate the coil's selfinductance. (b) If the current decreases uniformly from 5.00 A to 2.00 A in 3.00 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) \(L = 1.57 \times 10^{-3} \text{ H}\). (b) Self-induced EMF is \(-1.57 \text{ V}\). (c) EMF direction is from \(b\) to \(a\).

Step by step solution

01

Convert Units

First, convert all measurements into the appropriate SI units. Cross-sectional area, \(A = 6.25 \text{ cm}^2\), converts to \(6.25 \times 10^{-4} \text{ m}^2\). Mean radius, \(r = 4.00 \text{ cm}\), converts to \(0.04 \text{ m}\).
02

Calculate Cross-sectional Area and Length of Toroid

The cross-sectional area is already converted, but we need the length for the toroidal solenoid calculation. Using the mean radius, calculate the circumference, as the length, \(l = 2\pi r = 2 \pi \times 0.04 = 0.25133 \text{ m}\).
03

Compute Self-Inductance

The formula for the self-inductance \(L\) of a toroidal solenoid is \(L = \frac{\mu_0 N^2 A}{l}\). Plug in the known values: \(N = 500\), \(A = 6.25 \times 10^{-4} \text{ m}^2\), \(l = 0.25133 \text{ m}\), and \(\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}\). Thus, \[L = \frac{(4\pi \times 10^{-7}) \times 500^2 \times 6.25 \times 10^{-4}}{0.25133} \approx 1.57 \times 10^{-3} \text{ H}\].
04

Calculate Change in Current and Time

Calculate the change in current \(\Delta I\) and change in time \(\Delta t\). \(\Delta I = 5.00 \text{ A} - 2.00 \text{ A} = 3.00 \text{ A}\) and \(\Delta t = 3.00 \text{ ms} = 3.00 \times 10^{-3} \text{ s}\).
05

Calculate Self-Induced EMF

Use the formula for the self-induced EMF, \(\epsilon = -L \frac{\Delta I}{\Delta t}\). Substitute in known values: \(L = 1.57 \times 10^{-3} \text{ H}\), and calculated \(\Delta I = 3.00 \text{ A}\), \(\Delta t = 3.00 \times 10^{-3} \text{ s}\). \[ \epsilon = -(1.57 \times 10^{-3}) \cdot \frac{3.00}{3.00 \times 10^{-3}} = -1.57 \text{ V}\].
06

Determine Direction of Induced EMF

The negative sign in the calculated EMF indicates that the direction of the induced EMF is opposite to the direction of the current change. According to Lenz's Law, since the current reduces, the EMF induced will oppose the decrease. Therefore, if the current is flowing from \(a\) to \(b\), the EMF is induced from \(b\) to \(a\) to maintain the current.

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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 special type of coil that is wound in a donut-shaped form with a hollow center known as the torus. This design allows the magnetic field to be confined within the circular path, which is beneficial in applications that require a strong, consistent magnetic field without emissions to the outside. Unlike a straight solenoid, a toroidal solenoid does not have any poles because it is a continuous loop. This makes it very efficient as there is minimal leakage of the magnetic flux outside of the coils.
  • The windings are crucial as they determine the number of turns, which affects the self-inductance of the solenoid.
  • The cross-sectional area and length (or circumference) of the toroid are key parameters in calculating its inductance.
Understanding the construction and the magnetic field behavior in a toroidal solenoid helps improve the design and implementation of electronically inductive systems in a range of applications, from transformers to RF coils in MRI machines.
Self-Inductance
Self-inductance is a property inherent in coils, like the toroidal solenoid, where a change in current through the coil causes a change in the magnetic field, inducing a voltage (electromotive force or EMF) back into the coil itself. It is represented by the symbol \( L \) and its unit is the Henry (H).
The formula for self-inductance in a toroidal solenoid is given by: \[L = \frac{\mu_0 N^2 A}{l}\]where:
  • \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \text{ Tm/A} \).
  • \( N \) is the number of turns in the coil.
  • \( A \) is the cross-sectional area.
  • \( l \) is the mean length (circumference of the toroid).
Self-inductance reflects the ability of a coil to oppose changes in current, making it an important factor in the design of coils used in electrical circuits, including transformers and inductors.
Lenz's Law
Lenz's Law is a fundamental principle stating that an induced electromotive force (EMF) always acts to oppose the change in current that produced it. This is in accordance with the conservation of energy.
When the current in a coil changes, Lenz's Law determines the direction of the induced EMF. If the current decreases, as in the given example, the induced EMF will act to oppose this decrease by trying to maintain the current's original direction. Thus, the induced EMF works against fluctuations in current - a behavior central to the operation of inductors.
  • Helps in understanding the behavior of energy transfer in inductors.
  • Plays a critical role in AC circuits where the current is constantly changing.
This law is crucial in predicting the behavior of circuits experiencing variable currents, enabling the design of effective and efficient energy systems.
Self-Induced EMF
Self-induced EMF in a coil is the EMF generated within the same coil when the current flowing through it changes. The formula representing this phenomenon is: \[\epsilon = -L \frac{\Delta I}{\Delta t}\]where:
  • \( \epsilon \) is the self-induced EMF.
  • \( L \) is the self-inductance of the coil.
  • \( \Delta I \) is the change in current.
  • \( \Delta t \) is the change in time.
The negative sign follows from Lenz's Law, indicating that the induced EMF opposes the change in current. This opposition helps to stabilize the current flow and is a key aspect in the functioning of devices like transformers and motors. Understanding the principles of self-induced EMF aids engineers in designing circuits with the desired characteristics for specific applications, balancing the inductor's specifications with circuit performance requirements.

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Most popular questions from this chapter

An \(L\)-\(C\) circuit consists of a 60.0-mH inductor and a 250-\(\mu\)F capacitor. The initial charge on the capacitor is 6.00 \(\mu\)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?

A 7.50-nF capacitor is charged up to 12.0 V, then disconnected from the power supply and connected in series through a coil. The period of oscillation of the circuit is then measured to be 8.60 \(\times\) 10\(^{-5}\) s. Calculate: (a) the inductance of the coil; (b) the maximum charge on the capacitor; (c) the total energy of the circuit; (d) the maximum current in the circuit.

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

Two coils have mutual inductance \(M = 3.25 \times 10^{-4}\) H. The current \(i_1\) in the first coil increases at a uniform rate of 830 A/s. (a) What is the magnitude of the induced emf in the second coil? Is it constant? (b) Suppose that the current described is in the second coil rather than the first. What is the magnitude of the induced emf in the first coil?

(a) A long, straight solenoid has \(N\) turns, uniform cross sectional area \(A\), and length \(l\). Show that the inductance of this solenoid is given by the equation \(L = \mu_0 AN^2/l\). Assume that the magnetic field is uniform inside the solenoid and zero outside. (Your answer is approximate because \(B\) is actually smaller at the ends than at the center. For this reason, your answer is actually an upper limit on the inductance.) (b) A metallic laboratory spring is typically 5.00 cm long and 0.150 cm in diameter and has 50 coils. If you connect such a spring in an electric circuit, how much self-inductance must you include for it if you model it as an ideal solenoid?

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