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A small, closely wound coil has \(N\) turns, area \(A\), and resistance \(R\). The coil is initially in a uniform magnetic field that has magnitude \(B\) and a direction perpendicular to the plane of the loop. The coil is then rapidly pulled out of the field so that the flux through the coil is reduced to zero in time \(\Delta t\). (a) What are the magnitude of the average \(\operatorname{emf} \mathcal{E}_{\text {av }}\) and average current \(I_{\mathrm{av}}\) induced in the coil? (b) The total charge \(Q\) that flows through the coil is given by \(Q=I_{\mathrm{av}} \Delta t .\) Derive an expression for \(Q\) in terms of \(N, A, B,\) and \(R .\) Note that \(Q\) does not depend on \(\Delta t .\) (c) What is \(Q\) if \(N=150\) turns, \(A=4.50 \mathrm{~cm}^{2}, R=30.0 \Omega,\) and \(B=0.200 \mathrm{~T} ?\)

Short Answer

Expert verified
The total charge \(Q\) that flows through the coil is \(Q = 0.009 \, C\).

Step by step solution

01

Calculate Average EMF

First, the average emf needs to be calculated. From Faraday's law, the induced emf is given by the change in magnetic flux over time \(Δt\). The magnetic flux \(Φ\) is defined as the product of the magnetic field \(B\), the number of turns of wire \(N\), and the area of one loop \(A\). Hence, the magnetic flux change \(ΔΦ\) is \(NAB\). Therefore, \(\mathcal{E}_{\text {av }} = \frac{ΔΦ}{Δt} = \frac{NAB}{Δt}.\)
02

Calculate the Average current

Next, the average current can be calculated using Ohm's law, \(I = \frac{V}{R}\). Here, the voltage \(V\) is the induced emf \(\mathcal{E}_{\text {av }}\) and \(R\) is the resistance of the coil. So, \(I_{\mathrm{av}} = \frac{\mathcal{E}_{\text {av }}}{R} = \frac{NAB}{RΔt}.\)
03

Derive Expression for Charge

The total charge \(Q\) that flows through the coil can be found using the formula \(Q=I_{\mathrm{av}}Δt\). Thus, by substituting the value of \(I_{\mathrm{av}}\), we get \(Q = \frac{NAB}{R}.\) Note that the charge is independent of the time \(Δt\).
04

Calculate the Charge

For \(Q = NAB/R\), given that \(N=150\) turns, \(A=4.50 cm^2\), \(R=30.0 Ω\), and \(B=0.200 T\), we first need to convert the area from \(cm^2\) to \(m^2\) by multiplying by \(10^{-4}\) (since 1 \(cm^2 = 10^{-4} m^2\)). Therefore, \(A = 4.50 × 10^{-4} m^2\). Substituting these values into the formula, we find \(Q = \frac{150 × 0.200 × 4.50 × 10^{-4}}{30}.\)

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Key Concepts

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

Faraday's Law
Faraday's Law of Electromagnetic Induction is a fundamental principle in electromagnetism. It explains how an electromotive force (emf) is induced in a conductor when it experiences a change in magnetic flux. According to Faraday's Law,
  • The induced emf is directly proportional to the rate of change of magnetic flux.
  • The equation for Faraday's Law is given by: \( \mathcal{E} = -\frac{d\Phi}{dt} \)
Here, \( \mathcal{E} \) represents the induced emf and \( \Phi \) is the magnetic flux.

In the given exercise, when the coil is rapidly removed from the magnetic field, the magnetic flux becomes zero, causing a rapid change that induces an emf in the coil. This is quantified as the average emf using Faraday's Law formula:\[ \mathcal{E}_{\text{av}} = \frac{NAB}{\Delta t} \]where:
  • \( N \) is the number of turns in the coil
  • \( A \) is the area of each turn
  • \( B \) is the magnetic field strength
  • \( \Delta t \) is the time over which the change occurs
Magnetic Flux
Magnetic Flux is a measure of the quantity of magnetism, taking into account the strength and extent of a magnetic field. It is represented by the symbol \( \Phi \) and defined as the product of the magnetic field \( B \), the area \( A \) through which the field lines pass, and the cosine of the angle \( \theta \) between the magnetic field and the normal to the area.
  • The formula for magnetic flux is \( \Phi = NBA\cos\theta \).
  • In our exercise, because the magnetic field is perpendicular to the coil, \( \theta = 0 \), thus \( \cos 0 = 1 \), simplifying the flux to: \( \Phi = NAB \).
When the coil is removed from the field, the magnetic flux decreases from its initial value to zero, as the field no longer intersects the coil. This change in flux is what induces the emf, according to Faraday's Law.

To calculate the charge \( Q \), we only need to consider the initial magnetic flux. Since \( \Phi \) initially was \( NAB \) and becomes zero, the total change in flux \( \Delta \Phi \) is \( NAB \). Therefore, the relationship between charge and flux change is encapsulated in:\[ Q = \frac{NAB}{R} \]
Ohm's Law
Ohm's Law is a fundamental rule in electronics that relates the voltage, current, and resistance in an electrical circuit. It helps us understand how the quantities of an elastic conductor behave when carrying an electrical current. Ohm's Law is expressed by the formula:
  • \( I = \frac{V}{R} \)
Where:
  • \( I \) is the current in amperes (A)
  • \( V \) is the voltage in volts (V)
  • \( R \) is the resistance in ohms (\( \Omega \))
In the context of the coil from the exercise, the induced voltage, or emf, leads to an induced current \( I_{\text{av}} \).

According to Ohm's Law, this relationship is characterized as:\[ I_{\text{av}} = \frac{\mathcal{E}_{\text{av}}}{R} \]Given that the emf \( \mathcal{E}_{\text{av}} \) is \( \frac{NAB}{\Delta t} \), we can substitute this into Ohm's formula:\[ I_{\text{av}} = \frac{NAB}{R \Delta t} \]Finally, the step incorporates this current to calculate the total charge \( Q \) through:\[ Q = I_{\text{av}} \Delta t = \frac{NAB}{R} \]Thus, demonstrating how the average current and resulting charge depends on the coil's inherent characteristics and the changing magnetic field, while being independent of the time interval \( \Delta t \).

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Most popular questions from this chapter

Lightning Strikes. During lightning strikes from a cloud to the ground, currents as high as 25,000 A can occur and last for about \(40 \mu\) s. How much charge is transferred from the cloud to the earth during such a strike?

A typical small flashlight contains two batteries, each having an emf of \(1.5 \mathrm{~V},\) connected in series with a bulb having resistance \(17 \Omega .\) (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for \(5.0 \mathrm{~h}\), what is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.

A long, thin solenoid has 900 turns per meter and radius \(2.50 \mathrm{~cm} .\) The current in the solenoid is increasing at a uniform rate of \(36.0 \mathrm{~A} / \mathrm{s}\). What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) \(0.500 \mathrm{~cm}\) from the axis of the solenoid; (b) \(1.00 \mathrm{~cm}\) from the axis of the solenoid?

According to the U.S. National Electrical Code, copper wire used for interior wiring of houses, hotels, office buildings, and industrial plants is permitted to carry no more than a specified maximum amount of current. The table shows values of the maximum current \(I_{\max }\) for several common sizes of wire with varnished cambric insulation. The "wire gauge" is a standard used to describe the diameter of wires. Note that the larger the diameter of the wire, the smaller the wire gauge. $$ \begin{array}{ccc} \text { Wire gauge } & \text { Diameter }(\mathrm{cm}) & I_{\max }(\mathrm{A}) \\\ \hline 14 & 0.163 & 18 \\ 12 & 0.205 & 25 \\ 10 & 0.259 & 30 \\ 8 & 0.326 & 40 \\ 6 & 0.412 & 60 \\ 5 & 0.462 & 65 \\ 4 & 0.519 & 85 \end{array} $$ (a) What considerations determine the maximum current-carrying capacity of household wiring? (b) A total of \(4200 \mathrm{~W}\) of power is to be supplied through the wires of a house to the household electrical appliances. If the potential difference across the group of appliances is \(120 \mathrm{~V},\) determine the gauge of the thinnest permissible wire that can be used. (c) Suppose the wire used in this house is of the gauge found in part (b) and has total length \(42.0 \mathrm{~m}\). At what rate is energy dissipated in the wires? (d) The house is built in a community where the consumer cost of electrical energy is \(\$ 0.11\) per kilowatt-hour. If the house were built with wire of the next larger diameter than that found in part (b), what would be the savings in electricity costs in one year? Assume that the appliances are kept on for an average of 12 hours a day.

You apply a potential difference of \(4.50 \mathrm{~V}\) between the ends of a wire that is \(2.50 \mathrm{~m}\) in length and \(0.654 \mathrm{~mm}\) in radius. The resulting current through the wire is 17.6 A. What is the resistivity of the wire?

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