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The magnetic field within a long, straight solenoid with a eireular cross section and radius \(R\) is increasing at a rate of \(d B / d t .\) (a) What is the rate of change of flux through a circle with radius \(r_{1}\) inside the solenoid, normal to the axis of the solenoid, and with center on the solenoid axis? (b) Find the magnitude of the induced electric field inside the solenoid, at a distance \(r_{1}\) from its axis. Show the direction of this field in a diagram. (c) What is the magnitude of the induced electric field outside the solenoid, at a distance \(r_{2}\) from the axis? (d) Graph the magnitude of the induced electric field as a function of the distance \(r\) from the axis from \(r=0\) to \(r=2 R\) . (e) What is the magnitude of the induced emf in circular turn of radius \(R / 2\) that has its center on the solenoid axis? (f) What is the magnitude of the indnced emf if the radins in part (e) is \(R ?(g)\) What is the induced emf if the radius in part \((e)\) is 2\(R ?\)

Short Answer

Expert verified
(a) \(\pi r_1^2 \frac{dB}{dt}\); (b) \(\frac{1}{2} r_1 \frac{dB}{dt}\); (c) \(\frac{1}{2\pi r_2} R^2 \frac{dB}{dt}\); (d) graph as described; (e) \(\frac{\pi}{4} R^2 \frac{dB}{dt}\); (f)(g) \(\pi R^2 \frac{dB}{dt}\).

Step by step solution

01

Magnetic Flux Rate of Change Inside the Solenoid

The magnetic flux \( \Phi \) through a circle of radius \( r_1 \) inside the solenoid is given by \( \Phi = B \pi r_1^2 \). The rate of change of flux, \( \frac{d\Phi}{dt} \), due to the changing magnetic field \( B \), is \( \frac{d\Phi}{dt} = \frac{dB}{dt} \pi r_1^2 \). Thus, the rate of change of magnetic flux is \( \pi r_1^2 \frac{dB}{dt} \).
02

Induced Electric Field Inside the Solenoid

Using Faraday’s law, the induced emf \( \varepsilon \) in a circular loop is \( \varepsilon = - \frac{d\Phi}{dt} \). For a circle at distance \( r_1 \), the magnitude of the induced electric field \( E \) is given by \( E = \frac{\varepsilon}{2\pi r_1} = \frac{1}{2} r_1 \frac{dB}{dt} \). The direction of this field is tangential to the circle centered on the solenoid's axis.
03

Induced Electric Field Outside the Solenoid

Outside the solenoid, \( r_2 > R \), the magnetic field is approximately zero. Thus, the induced electric field \( E \) at this distance is calculated by considering the full solenoid surface area's changing flux: \( E = \frac{1}{2\pi r_2} R^2 \frac{dB}{dt} \). This field circulates around the solenoid.
04

Graph of Induced Electric Field Magnitude

The graph plotting \( E \) against \( r \) from \( r = 0 \) to \( r = 2R \) is linear inside the solenoid (\( 0 \leq r \leq R \)) and decreases inversely with \( r \) outside (\( r \geq R \)), since the field is zero right outside the solenoid and increases negatively as \( r \) grows.
05

Induced EMF for Radius \( R/2 \)

For a circular turn of radius \( R/2 \) with its center on the solenoid axis, the induced emf\( \varepsilon_{\frac{R}{2}} \) is \( \frac{\pi (R/2)^2 dB/dt}{dt} = \frac{\pi R^2}{4} \frac{d B}{d t} \).
06

Induced EMF for Radius \( R \)

For a circular turn of radius \( R \), the induced emf is \( \varepsilon_{R} = \pi R^2 \frac{d B}{d t} \).
07

Induced EMF for Radius \( 2R \)

For a turn of radius \( 2R \), the emf includes full flux change over the solenoid's area to \( R \): \( \varepsilon_{2R} = \pi R^2 \frac{d B}{d t} \). External radius does not count beyond \( R \).

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Key Concepts

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

Magnetic Flux
Magnetic flux, denoted by the symbol \( \Phi \), is a measure of the quantity of magnetic field passing through a given area. It is especially important when examining the behavior of magnetic fields in closed loops like those found inside a solenoid. To calculate magnetic flux through a loop, we use the formula \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field strength, \( A \) is the area of the loop, and \( \theta \) is the angle between the magnetic field lines and the perpendicular to the surface. In the context of a solenoid, where the cross-section is circular, the angle \( \theta \) is usually zero, making the cosine factor equal to 1. Thus, the expression simplifies to \( \Phi = B \cdot A \). For a circle inside the solenoid, \( A = \pi r_1^2 \), where \( r_1 \) is the radius of the circular area of interest. Understanding the rate of change of this flux is essential for determining the induced effects like voltage and electric fields.
Faraday's Law
Faraday's Law of electromagnetic induction is a fundamental principle which describes how electric currents can be induced by changing magnetic environments. It states that the induced electromotive force (emf) in any closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, this is expressed as \( \varepsilon = - \frac{d\Phi}{dt} \). This means that the induced emf is proportional to how quickly the magnetic flux is changing. The negative sign is a consequence of Lenz's Law, ensuring that the induced current will oppose the change in flux. In the context of a solenoid, the changing magnetic field results in a change in magnetic flux, leading to an induced emf along a loop inside or around the solenoid. This is the core mechanism behind the generation of electric fields in and around coils and solenoids.
Induced Electric Field
An induced electric field is generated when there is a change in magnetic flux according to Faraday's Law. For a solenoid, as the magnetic field inside the solenoid changes, it induces an electric field in the surrounding space. This induced field is often expressed in terms of its magnitude \( E \) and its direction. Inside a solenoid, at a distance \( r_1 \) from its axis, the induced electric field can be calculated using the formula \( E = \frac{1}{2} r_1 \frac{dB}{dt} \). This indicates that the induced electric field is directly proportional to both the rate of change of the magnetic field \( \frac{dB}{dt} \) and the distance \( r_1 \) from the solenoid's center. The direction of this field is tangential to the circle created by the distance \( r_1 \). Conversely, outside the solenoid, the electric field behaves differently. Despite the magnetic field being practically zero outside, changes in magnetic flux still induce an electric field, circulating around the solenoid.
Solenoid
A solenoid is a coil of wire designed to generate a magnetic field when an electric current flows through it. They are commonly used in various applications to create controlled magnetic fields. Inside a solenoid, the magnetic field is strong and uniform, while outside, it is weak and quickly dissipates. The strength of the field inside a solenoid can be calculated by the expression \( B = \mu_0 n I \), where \( \mu_0 \) is the permeability of free space, \( n \) is the number of turns per unit length, and \( I \) is the current flowing through the coil. The uniform nature of the magnetic field inside the solenoid makes it an excellent tool for experiments and applications involving electromagnetic induction. When the current changes, the magnetic field also changes, leading to the phenomena explained by Faraday's Law and the resulting induced electric fields both inside and encircling the solenoid. This characteristic makes solenoids crucial components in technologies such as electrical transformers, inductors, and sensors.

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Most popular questions from this chapter

A closely wound search coil (Exercise 29.3) has an area of \(3.20 \mathrm{cm}^{2}, 120\) turns, and a resistance of \(60.0 \Omega .\) It is connected to a charge-measuring instrument whose resistance is 45.0\(\Omega\) . When the coil is rotated quickly from a position parallel to a uniform magnetic field to a position perpendicular to the field, the instrument indicates a charge of \(3.56 \times 10^{-5} \mathrm{C}\) . What is the magnitude of the field?

A capacitor has two parallel plates with area \(A\) separated by a distance \(d\) . The space between plates is filled with a material having dielectric constant \(K\) . The material is not a perfect insulator but has resistivity \(\rho\) . The capacitor is initially charged with charge of magnitude \(Q_{0}\) on each plate that gradually discharges by conduction through the dielectric. (a) Calculate the conduction current density \(j_{\mathrm{C}}(t)\) in the dielectric. (b) Show that at any instant the dis-placement current density in the diclectric is equal in magnitude to the oonduotion current density but opposite in direction, so the total current density is zero at every instant.

A long, straight wire made of a type-I superconductor carries a constant current \(I\) along its length. Show that the current cannot be uniformly spread over the wire's cross section but instead must all be at the surface.

A circular wire loop of radius \(a\) and resistance \(R\) initially has a magnetic flux through it due to an external magnetic field. The extermal field then decreases to zero. A current is induced in the loop while the external field is changing; however, this current does not stop at the instant that the external field stops changing. The reason is that the current itself generates a magnetic field, which gives rise to a flux through the loop. If the current changes, the flux through the loop changes as well, and an induced emf appears in the loop to oppose the change. (a) The magnetic field at the center of the loop of radius a produced by a current i in the loop is given by \(B=\mu_{0} i / 2 a\) . If we use the crode approximation that the field has this same value at all points within the loop, what is the flux of this field through the loop? (b) By using Faraday's law, Eq. \((29.3),\) and the relationship \(\mathcal{E}=i R,\) show that after the external field has stopped changing, the current in the loop obeys the differential equation $$ \frac{d i}{d t}=-\left(\frac{2 R}{\pi \mu_{0} a}\right) i $$ (c) If the current has the value \(i_{0}\) at \(t=0\) , the instant that the external field stops changing. snive the equation in part \((b)\) to find \(i\) as a function of time for \(t>0 .\) (Hint: In Section 26.4 we encountered a similar differential equation, Eq. \((26.15),\) for the quantity \(q .\) This equation for \(i\) may be solved in the same way. (d) If the loop has radius \(a=50 \mathrm{cm}\) and resistance \(R=0.10 \Omega,\) how long after the external field stops changing will the current be equal to 0.010 \(\mathrm{o}\) (that is, \(\frac{1}{100}\) of its initial value)? (e) In solving the examples in this chapter, we ignored the effects described in this problem. Explain why this is a good approximation.

emf in a Bullet. At the equator, the earth's magnetic field is approximately horizontal, is directed towand the north, and has a value of \(8 \times 10^{-5} \mathrm{T}\) . (a) Estimate the emf induced between the top and bottom of a bullet shot horizontally at a target on the equator if the bullet is shot toward the east. Assume the bullet has a length of 1 \(\mathrm{cm}\) and a diamcter of 0.4 \(\mathrm{cm}\) and is traveling at 300 \(\mathrm{m} / \mathrm{s}\) . Which is at higher potential: the top or boutom of the bullet? (b) What is the emfif the bullet travels south?(c) What is the emf induced between the front and back of the bullet for any horizontal velocity?

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