/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 The magnetic field within a long... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The magnetic field within a long, straight solenoid with a circular 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 a circular turn of radius \(R / 2\) that has its center on the solenoid axis? (f) What is the magnitude of the induced emf if the radius in part (e) is \(R ?\) (g) What is the induced emf if the radius in part (e) is 2R?

Short Answer

Expert verified
(a) \( \pi r_1^2 \frac{dB}{dt} \); (b) \( \frac{r_1}{2} \frac{dB}{dt} \) (circumferential); (c) \( \frac{R^2}{2r_2} \frac{dB}{dt} \); (d) Follow the provided description; (e) \( \pi \frac{R^2}{4} \frac{dB}{dt} \); (f) \( \pi \frac{R^2}{2} \frac{dB}{dt} \); (g) \( \frac{3\pi R^2}{2} \frac{dB}{dt} \).

Step by step solution

01

Calculate Rate of Change of Magnetic Flux

The magnetic flux through the circle of radius \( r_1 \) within the solenoid is given by \( \Phi = B \cdot A = B \cdot \pi r_1^2 \), where \( A = \pi r_1^2 \) is the area of the circle. The rate of change of this flux is \( \frac{d\Phi}{dt} = \pi r_1^2 \cdot \frac{dB}{dt} \).
02

Induced Electric Field Inside the Solenoid

Using Faraday's Law, the magnitude of the induced electric field \( E \) at a distance \( r_1 \) from the solenoid's axis is given by \( E \cdot 2\pi r_1 = - \frac{d\Phi}{dt} \). Substituting the expression for \( \frac{d\Phi}{dt} \), we get \( E = \frac{\pi r_1^2 \frac{dB}{dt}}{2 \pi r_1} = \frac{r_1}{2} \frac{dB}{dt} \). The direction of the induced electric field is circumferential around the axis, using the right-hand rule.
03

Induced Electric Field Outside the Solenoid

For \( r_2 > R \), the enclosed magnetic flux by a circle of radius \( r_2 \) is still = \( B \pi R^2 \), as only the magnetic field inside the solenoid changes. From Faraday’s Law: \( E \cdot 2\pi r_2 = - \frac{d(B \pi R^2)}{dt} \), so \( E = \frac{R^2}{2r_2} \frac{dB}{dt} \).
04

Graph Magnitude of Induced Electric Field as a Function of Distance

For \( r \leq R \), the induced field \( E = \frac{r}{2} \frac{dB}{dt} \). For \( r > R \), the induced field \( E = \frac{R^2}{2r} \frac{dB}{dt} \). Plotting \( E \) against \( r \), we see a linear increase inside the solenoid up to \( R \), after which the field decreases with \( \frac{1}{r} \).
05

Calculate Induced EMF for Radius R/2

The induced emf \( \mathcal{E} \) in a loop of radius \( R/2 \) inside the solenoid is given by \( \mathcal{E} = E \cdot 2\pi (R/2) = \pi \frac{R^2}{4} \frac{dB}{dt} \).
06

Calculate Induced EMF for Radius R

The induced emf for a loop of radius \( R \) (inside the solenoid) is \( \mathcal{E} = \pi \frac{R^2}{2} \frac{dB}{dt} \).
07

Calculate Induced EMF for Radius 2R

For a loop with radius \( 2R \), which is partially outside the solenoid, the emf is \( \mathcal{E} = \left(\frac{R^2}{2} + R^2 \right) \pi \frac{dB}{dt} = \frac{3\pi R^2}{2} \frac{dB}{dt} \), summing contributions inside and outside.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 that describes how electric fields can be generated by changing magnetic fields. This law is the cornerstone of understanding how electromagnetic fields interact. It states that the electromotive force (EMF) induced in a closed loop is proportional to the rate of change of magnetic flux through the loop. Mathematically, it's expressed as \( \mathcal{E} = -\frac{d\Phi}{dt} \), where \( \mathcal{E} \) is the induced EMF and \( \Phi \) is the magnetic flux through the loop.

When the magnetic field around a loop changes—either by varying the field intensity, moving the magnet, or changing the loop's orientation—an electric current is induced. This principle forms the basis of many electrical devices, such as transformers, electric generators, and inductors.
  • Induced EMF depends on the rate of change of flux, not the flux itself.
  • The negative sign in Faraday’s law equation signifies Lenz's law, indicating that the induced EMF creates a current and magnetic field that opposes the initial change.
  • This law can be applied to loops of various shapes and in different magnetic fields.
Magnetic Flux
Magnetic flux is essentially a measure of the number of magnetic field lines passing through a given area. It's a concept that helps us understand the field's influence over different surfaces. The magnetic flux \( \Phi \) is calculated as \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field strength, \( A \) is the area through which the lines pass, and \( \theta \) is the angle between the magnetic field lines and the perpendicular to the surface.

Changes in magnetic flux are integral to the production of electric current, using the principle of electromagnetic induction.
  • Flux is maximum when the field lines are perpendicular to the surface (i.e., \( \theta = 0\)).
  • If the loop is tilted such that field lines are parallel, flux is zero because \( \cos(90^\circ) = 0 \).
  • Flux changes can occur by varying the field strength, the area, or the orientation of the surface.
Solenoid
A solenoid is a coil of wire designed to create a uniform magnetic field in its interior space when an electric current is passed through it. This device is central to many electromagnetic applications. The field lines in a solenoid are parallel and closely spaced, creating a strong, consistent field at its core while rapidly diminishing outside.

The strength of the magnetic field inside a solenoid is given by \( B = \mu n I \), where \( \mu \) is the magnetic permeability of the core, \( n \) is the number of turns per unit length, and \( I \) is the current passing through the wire.
  • Fields are strong inside due to the alignment and density of field lines.
  • Outside the solenoid, the field is much weaker and tends to cancel out due to opposite loops of current.
  • Solenoids are used in applications like electromagnets, valves, and loops to induce specific field patterns.
Induced Electric Field
An induced electric field is created as a result of changing magnetic flux, as per Faraday's Law. This field acts tangentially along any closed loop surrounding the changing magnetic field. When considering a solenoid, the varying magnetic field within induces an electric field that forms concentric loops around the solenoid's axis.

The magnitude of this induced field inside a solenoid at a distance \( r \) from its axis is determined using the equation \( E = \frac{r}{2} \frac{dB}{dt} \). For a position outside the solenoid, where \( r > R \), it is calculated using \( E = \frac{R^2}{2r} \frac{dB}{dt} \).
  • The field lines of the induced electric field form closed loops, encircling the changing magnetic field lines.
  • This field is non-conservative, meaning the work done in moving a charge around the loop isn't zero as it would be in a static field.
  • The induced current direction follows the right-hand rule, aligning with the field to always oppose changes in flux, as explained by Lenz's Law.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could operate using a generator but you have no magnets. The earth's magnetic field at your location is horizontal and has magnitude \(8.0 \times 10^{-5} \mathrm{T},\) and you decide to try to use this field for a generator by rotating a large circular coil of wire at a high rate. You need to produce a peak emf of 9.0 \(\mathrm{V}\) and estimate that you can rotate the coil at 30 \(\mathrm{rpm}\) by turning a crank handle. You also decide that to have an acceptable coil resistance, the maximum number of turns the coil can have is 2000 . (a) What area must the coil have? (b) If the coil is circular, what is the maximum translational speed of a point on the coil as it rotates? Do you think this device is feasible? Explain.

A circular conducting ring with radius \(r_{0}=0.0420 \mathrm{m}\) lies in the \(x y\) -plane in a region of uniform magnetic field \(\vec{B}=B_{0}\left[1-3\left(t / t_{0}\right)^{2}+\right.\) 2\(\left(t / t_{0}\right)^{3} \hat{\boldsymbol{k}} .\) In this expression, \(t_{0}=\) 0.0100 \(\mathrm{s}\) and is constant, \(t\) is time, \(\hat{\boldsymbol{k}}\) is the unit vector in the \(+z\) - direction, and \(B_{0}=0.0800 \mathrm{T}\) and is constant. At points \(a\) and \(b(\) Fig. \(\mathrm{P} 29.64)\) there is a small gap in the ring with wires leading to an external circuit of resistance \(R=12.0 \Omega .\) There is no magnetic field at the location of the external circuit. (a) Derive an expression, as a function of time, for the total magnetic flux \(\Phi_{B}\) through the ring. (b) Determine the emf induced in the ring at time \(t=5.00 \times 10^{-3}\) s. What is the polarity of the emf? (c) Because of the internal resistance of the ring, the current through \(R\) at the time given in part (b) is only 3.00 mA. Determine the internal resistance of the ring. (d) Determine the emf in the ring at a time \(t=1.21 \times 10^{-2}\) s. What is the polarity of the emf? (e) Determine the time at which the current through \(R\) reverses its direction.

CALC In a region of space, a magnetic field points in the \(+x\) -direction (toward the right). Its magnitude varies with position according to the formula \(B_{x}=B_{0}+b x,\) where \(B_{0}\) and \(b\) are positive constants, for \(x \geq 0 .\) A flat coil of area \(A\) moves with uniform speed \(v\) from right to left with the plane of its area always perpendicular to this field. (a) What is the emf induced in this coil while it is to the right of the origin? (b) As viewed from the origin, what is the direction (clockwise or counterclockwise) of the current induced in the coil? (c) If instead the coil moved from left to right, what would be the answers to parts (a) and (b)?

A long, thin solenoid has 400 turns per meter and radius 1.10 \(\mathrm{cm} .\) The current in the solenoid is increasing at a uniform rate \(d i / d t .\) The induced electric field at a point near the center of the solenoid and 3.50 \(\mathrm{cm}\) from its axis is \(8.00 \times 10^{-6} \mathrm{V} / \mathrm{m}\) . Calculate di/dt.

Are Motional emfs a Practical Source of Electricity? How fast (in \(\mathrm{m} / \mathrm{s}\) and mph) would a 5.00 -cm copper bar have to move at right angles to a \(0.650-\mathrm{T}\) magnetic field to generate 1.50 \(\mathrm{V}\) (the same as a A \(\mathrm{A}\) battery) across its ends? Does this seem like a practical way to generate electricity?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.