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Chapter 30: Problem 14

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 self-induced emf to equal 7.50 \(\mathrm{mV} ?\)

Short Answer

Expert verified
The rate of change of current is \(8.36 \times 10^{-3}\) A/s.

Step by step solution

01

Understanding the Problem

We need to find the rate of change of the current in a solenoid, given some initial conditions, so that the self-induced emf is 7.50 mV. We know the number of turns, the current, and the magnetic flux.
02

Using the Formula for Self-Induced EMF

Self-induced emf in a solenoid is given by the formula:\[\mathcal{E} = -L \frac{di}{dt}\]where \(\mathcal{E}\) is the emf, \(L\) is the inductance, and \(\frac{di}{dt}\) is the rate of change of current. We need to express \(\frac{di}{dt}\) in terms of known quantities.
03

Calculating Inductance of Solenoid

Inductance \(L\) of a solenoid is given by:\[L = \frac{N \Phi}{I}\]where \(N\) is the number of turns, \(\Phi\) is the magnetic flux per turn, and \(I\) is the current. Plugging in given values: \(N = 800\), \(\Phi = 3.25 \times 10^{-3}\) Wb, \(I = 2.90\) A.
04

Substituting Values into Inductance Formula

Calculate \(L\):\[L = \frac{800 \times 3.25 \times 10^{-3}}{2.90} = 0.8966 \text{ H}\]
05

Rearranging EMF Formula for Rate of Change of Current

We need to find \(\frac{di}{dt}\):\[7.50 \times 10^{-3} = -0.8966 \cdot \frac{di}{dt}\]Solving for \(\frac{di}{dt}\):\[\frac{di}{dt} = \frac{-7.50 \times 10^{-3}}{-0.8966} = 8.36 \times 10^{-3} \text{ A/s}\]
06

Final Answer

The magnitude of the rate of change of the current \(\left|\frac{di}{dt}\right|\) must be \(8.36 \times 10^{-3}\) A/s.

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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 in a cylindrical shape. When electric current passes through it, the solenoid generates a magnetic field similar to that of a bar magnet. The strength of this magnetic field is directly proportional to the number of turns in the solenoid and the current flowing through it. Solenoids are widely used because they can create controlled magnetic fields of varying strength by adjusting the current. In practical applications, you will find solenoids in devices like electromagnetic locks and valves.
  • A solenoid becomes stronger with more turns or increased current.
  • It forms the fundamental structure for electromagnets.
  • Is utilized in various electronic applications.
Self-Induced EMF
Self-induced electromotive force (emf) occurs when a change in current through a coil induces a voltage (emf) in the same coil. This phenomenon primarily results from a changing magnetic field produced by the current change in the coil.The induced emf opposes the change in current, according to Lenz's law. Mathematically, it is represented by \[ \mathcal{E} = -L \frac{di}{dt} \]where \( \mathcal{E} \) is the emf, \( L \) is the inductance, and \( \frac{di}{dt} \) is the rate of change of current. This self-induction property is crucial for understanding how circuits and electromagnetic devices operate.
  • Causes a voltage to oppose current change.
  • Fundamental in alternating current (AC) circuits.
  • Makes transformers and inductors possible.
Magnetic Flux
Magnetic flux captures the amount of magnetic field passing through a given area. It is an essential concept for understanding electromagnetic principles. In the context of a solenoid, it refers to the magnetic field passing through each turn of the coil.The formula of magnetic flux \( \Phi \) is expressed as:\[ \Phi = B \cdot A \]where \( B \) is the magnetic field strength and \( A \) is the area it penetrates.
  • Measured in Webers (Wb).
  • Indicates field strength and area coverage.
  • Vital for calculating induced currents and voltages.
Inductance
Inductance is the property of a coil that quantifies its ability to induce an emf as a result of changes in current. This effect results from the magnetic field interactions within the coil.The inductance \( L \) of a solenoid is given by:\[ L = \frac{N \Phi}{I} \]where \( N \) is the number of turns, \( \Phi \) is the average magnetic flux through each turn, and \( I \) is the current.With this relationship, inductance becomes a measure of a solenoid's capability to oppose changes in current. It is crucial for the operation of many electrical devices such as transformers and inductors.
  • Measured in Henry (H).
  • Determines the magnetic energy storage potential.
  • Essential in filters and tuning circuits.

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Most popular questions from this chapter

A toroidal solenoid with mean radius \(r\) and cross-sectional area \(A\) is wound uniformly with \(N_{1}\) turns. A second toroidal solenoid with \(N_{2}\) turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius. (a) What is the mutual inductance of the two solenoids? Assume that the magnetic field of the first solenoid is uniform across the cross section of the two solenoids. (b) If \(N_{1}=500\) turns, \(N_{2}=300\) turns, \(r=10.0 \mathrm{cm},\) and \(A=0.800 \mathrm{cm}^{2},\) what is the value of the mutual inductance?

Solar Magnetic Energy. Magnetic fields within a sunspot can be as strong as 0.4 \(\mathrm{T}\) . (By comparison, the earth's magnetic field is about \(1 / 10,000\) as strong.) Sunspots can be as large as \(25,000 \mathrm{km}\) in radius. The material in a sunspot has a density of about \(3 \times 10^{-4} \mathrm{kg} / \mathrm{m}^{3}\) . Assume \(\mu\) for the sunspot material is \(\mu_{0}\) . If 100\(\%\) of the magnetic-field energy stored in a sunspot could be used to eject the sunspot's material away from the sun's surface, at what speed would that material be ejected? Compare to the sun's escape speed, which is about \(6 \times 10^{5} \mathrm{m} / \mathrm{s} .\) Hint: Calculate the kinetic energy the magnetic field could supply to 1 \(\mathrm{m}^{3}\) of sunspot material.)

An inductor with an inductance of 2.50 \(\mathrm{H}\) and a resistance of 8.00\(\Omega\) is connected to the terminals of a battery with an emf of 6.00 \(\mathrm{V}\) and negligible internal resistance. Find (a) the initial rate of increase of current in the circuit; (b) the rate of increase of current at the instant when the current; is \(0.500 \mathrm{A} ;\) (c) the current 0.250 \(\mathrm{s}\) after the circuit is closed; (d) the final steady-state current.

An inductor used in a dc power supply has an inductance of 12.0 \(\mathrm{H}\) and a resistance of 180\(\Omega .\) It carries a current of 0.300 \(\mathrm{A}\) . (a) What is the energy stored in the magnetic field? (b) At what rate is thermal energy developed in the inductor? (c) Does your answer to part (b) mean that the magnetic-field energy is decreasing with time? Explain.

An \(L-C\) circuit containing an \(80.0-\mathrm{mH}\) inductor and a \(1.25-\mathrm{nF}\) capacitor oscillates with a maximum current of 0.750 \(\mathrm{A}\) . Calculate: (a) the maximum charge on the capacitor and (b) the oscillation frequency of the circuit. (c) Assuming the capacitor had its maximum charge at time \(t=0\) , calculate the energy stored in the inductor after 2.50 \(\mathrm{ms}\) of oscillate the energy stored in.
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