/*! 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 1 A flat, rectangular coil consist... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 29: Problem 1

A flat, rectangular coil consisting of 50 tums measures 25.0 \(\mathrm{cm}\) by 30.0 \(\mathrm{cm}\) . It is in a uniform, \(1.20-\mathrm{T}\) , magnetic field, with the plane of the coil parallel to the field. In 0.222 s, it is rotated so that the plane of the coil is perpendicular to the field. (a) What is the change in the magnetic flux through the coil due to this rotation? (b) Find the magnitude of the average emf induced in the coil during this rotation.

Short Answer

Expert verified
(a) Change in flux: -9.00 \(\mathrm{Wb}\); (b) Average EMF: 2.03 \(\mathrm{V}\).

Step by step solution

01

Calculate Initial Magnetic Flux

The initial magnetic flux \( \Phi_i \) is given by \( \Phi_i = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field strength, \( A \) is the area of the coil, and \( \theta \) is the angle between the field and the normal to the coil's surface. Since the plane of the coil is initially parallel to the field, \( \theta = 0^\circ \), so \( \cos(\theta) = 1 \). Thus, \( \Phi_i = 1.20 \times (0.25 \times 0.30) \). Calculate \( \Phi_i \).
02

Calculate Final Magnetic Flux

Once the coil is rotated to be perpendicular to the field, the angle \( \theta \) between the field and the normal to the coil becomes \( 90^\circ \), so \( \cos(\theta) = 0 \). Thus, the final magnetic flux \( \Phi_f = B \cdot A \cdot 0 = 0 \).
03

Calculate Change in Magnetic Flux

The change in magnetic flux \( \Delta \Phi \) is given by \( \Delta \Phi = \Phi_f - \Phi_i \). Substitute the values of \( \Phi_f \) and \( \Phi_i \) to find \( \Delta \Phi \).
04

Calculate Area of the Coil

The area \( A \) of the coil is the product of its length and width.\( A = 0.25 \times 0.30 \; \text{m}^2 \). Calculate this value and use it to compute the initial flux in Step 1.
05

Compute Induced EMF

According to Faraday's Law, the magnitude of the average induced EMF is \( |\mathcal{E}| = \frac{N \times \Delta \Phi}{\Delta t} \), where \( N \) is the number of turns, and \( \Delta t \) is the time period. Substitute the given values: \( N = 50 \), \( \Delta t = 0.222 \; \text{s} \), and your computed \( \Delta \Phi \) from Step 3 to find \( |\mathcal{E}| \).

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 91影视!

Key Concepts

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

Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, taking into account the strength and the extent of a magnetic field. It basically describes how much magnetic field passes through a particular area.
The formula to calculate magnetic flux \( \Phi \) is given by:
  • \( \Phi = B \times A \times \cos(\theta) \)
where:
  • \( B \) is the magnetic field strength,
  • \( A \) is the area through which the field lines pass,
  • \( \theta \) is the angle between the field lines and the perpendicular (normal) to the surface of the coil.
For a coil that is parallel to the field, \( \theta = 0^\circ \) and \( \cos(\theta) = 1 \), meaning all the magnetic field lines pass through the coil. When it's perpendicular, \( \theta = 90^\circ \) and \( \cos(\theta) = 0 \), thus, no field lines pass through. In the exercise, changing from parallel to perpendicular results in a significant change in magnetic flux, which is essential for inducing an electromotive force (EMF) in a coil.
Faraday's Law of Induction
Faraday's Law of Induction helps explain how electric currents are generated by changing magnetic fields. Essentially, this law states that the induced electromotive force (EMF) in a closed circuit is directly proportional to the rate of change of magnetic flux through the circuit. This relationship can be written as:
  • \( \mathcal{E} = -\frac{d\Phi}{dt} \)
The negative sign in the equation is due to Lenz's Law, which tells us that the induced EMF will always work in a direction to oppose the change in flux that produced it.
This principle is fundamental in understanding why rotating the rectangular coil in the textbook exercise induces an EMF. The coil starts parallel to a magnetic field and then becomes perpendicular, creating a change in magnetic flux from maximum to zero, and therefore, according to Faraday's Law, an average EMF is induced over the rotation period.
Rectangular Coil
A rectangular coil, such as the one described in the exercise, is often used in laboratories and experimental setups to demonstrate magnetic effects and calculate induced EMF. The typical rectangular coil consists of several turns of wire wound in a rectangular shape.
Key characteristics are:
  • Area (A): Product of the coil's length and width, affecting magnetic flux. Here it's calculated as \( 0.25 \times 0.30 \; \text{m}^2 \), providing a surface for the field to interact with.
  • Number of Turns (N): More turns mean higher induced EMF for a given change in flux, as they effectively multiply the effect of a single loop. The exercise specifies 50 turns.
When the coil rotates from being parallel to perpendicular to the magnetic field, the rapid change in field interaction results in a change in the magnetic flux and, therefore, an induced EMF. This setup helps students understand how real-world applications like electric generators and transformers operate by similar mechanisms.

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

Shrinking Loop. A circular loop of flexible iron wire has an initial circunference of \(165.0 \mathrm{cm},\) but its circunference is decreasing at a constant rate of 12.0 \(\mathrm{cm} / \mathrm{s}\) due to a tangential pull on the wire. The loop is in a constant, viniform magnetic field oriented perpendicular to the plane of the loop and with magnitude 0.500 \(\mathrm{T}\) . (a) Find the emf induced in the loop at the instant when 9.0 s have passed. (b) Find the direction of the induced current in the loop as viewed looking along the direction of the magnetic field.

A metal ring 4.50 \(\mathrm{cm}\) in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic ficld. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 \(\mathrm{T} / \mathrm{s}\) (a) What is the magnitude of the electric field induced in the ring? (b) In which direction (clockwise or counterclockwise) does the current flow as viewed by someone on the south pole of the magnet?

Make a Generator? You are shipwrecked on a deserted tropical island. You have some electrical devices that you could uperate using a generalor but you have nu maguels. The eardis 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 mumber 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 this device is feasible? Explain.

Displacement Current in a Wire. A long, straight, copper wire with a circular cross-scctional area of 2.1 \(\mathrm{mm}^{2}\) carries a current of 16 \(\mathrm{A}\) . The resistivity of the material is \(20 \times 10^{-8} \Omega \cdot \mathrm{m}\) . (a) What is the uniform electric field in the material? (b) If the cur- rent is changing at the rate of 4000 \(\mathrm{A} / \mathrm{s}\) , at what rate is the electric field in the material changing? (c) What is the displacement current density in the material in part (b)? (Hint: Since \(K\) for copper is very close to \(1,\) use \(\epsilon=\epsilon_{0} . )\) (d) If the current is changing as in part (b), what is the magnitude of the magnetic field 6.0 \(\mathrm{cm}\) from the center of the wire? Note that both the conduction current and the displacement current should be included in the calculation of \(B\) . Is the contribution from the displacement current significant?

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.
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Scientific Method Physics

Read Explanation

Quantum Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
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.