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Chapter 25: Problem 4

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 \(螖t\). The magnetic flux \(桅\) 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 \(螖桅\) is \(NAB\). Therefore, \(\mathcal{E}_{\text {av }} = \frac{螖桅}{螖t} = \frac{NAB}{螖t}.\)
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, \(I_{\mathrm{av}} = \frac{\mathcal{E}_{\text {av }}}{R} = \frac{NAB}{R螖t}.\)
03

Derive Expression for Charge

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

Calculate the Charge

For \(Q = NAB/R\), given that \(N=150\) turns, \(A=4.50 cm^2\), \(R=30.0 惟\), 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, \(A = 4.50 脳 10^{-4} m^2\). Substituting these values into the formula, we find \(Q = \frac{150 脳 0.200 脳 4.50 脳 10^{-4}}{30}.\)

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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

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

BIO The average bulk resistivity of the human body (apart from surface resistance of the skin) is about \(5.0 \Omega \cdot \mathrm{m}\). The conducting path between the hands can be represented approximately as a cylinder \(1.6 \mathrm{~m}\) long and \(0.10 \mathrm{~m}\) in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) What potential difference between the hands is needed for a lethal shock current of \(100 \mathrm{~mA}\) ? (Note that your result shows that small potential differences produce dangerous currents when the skin is damp.) (c) With the current in part (b), what power is dissipated in the body?

Current passes through a solution of sodium chloride. In \(1.00 \mathrm{~s}, 2.68 \times 10^{16} \mathrm{Na}^{+}\) ions arrive at the negative electrode and \(3.92 \times 10^{16} \mathrm{Cl}^{-}\) ions arrive at the positive electrode. (a) What is the current passing between the electrodes? (b) What is the direction of the current?

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?

BIO Electric Eels. Electric eels generate electric pulses along their skin that can be used to stun an enemy when they come into contact with it. Tests have shown that these pulses can be up to \(500 \mathrm{~V}\) and produce currents of \(80 \mathrm{~mA}\) (or even larger). A typical pulse lasts for \(10 \mathrm{~ms}\). What power and how much energy are delivered to the unfortunate enemy with a single pulse, assuming a steady current?
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