/*! 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 12 A horizontal rectangular surface... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 12

A horizontal rectangular surface has dimensions 2.80 \(\mathrm{cm}\) by 3.20 \(\mathrm{cm}\) and is in a uniform magnetic field that is directed at an angle of \(30.0^{\circ}\) above the horizontal. What must the magnitude of the magnetic field be in order to produce a flux of \(4.20 \times 10^{-4} \mathrm{Wb}\) through the surface?

Short Answer

Expert verified
The magnetic field magnitude must be approximately 0.542 T.

Step by step solution

01

Determine Surface Area

First, calculate the surface area (A) of the rectangle using the formula for the area of a rectangle, which is width times height. Given dimensions are: width = 2.80 cm and height = 3.20 cm, so \( A = 2.80 \times 3.20 \ \mathrm{cm^2}\). Convert to square meters by using \(1 \ \mathrm{cm^2} = 10^{-4} \ \mathrm{m^2}\) to get \(A = 8.96 \times 10^{-4} \ \mathrm{m^2}\).
02

Apply Magnetic Flux Formula

Magnetic flux (\(\Phi\)) is given by the formula \(\Phi = B \cdot A \cdot \cos(\theta)\), where \(B\) is the magnetic field magnitude, \(A\) is the area, and \(\theta\) is the angle between the magnetic field and the normal to the surface. Here, \(\theta = 30.0^{\circ}\), \(\cos(30.0^{\circ}) = \sqrt{3}/2\).
03

Rearrange Formula to Solve for Magnetic Field

To find the magnetic field magnitude \(B\), rearrange the magnetic flux formula to solve for \(B\): \(B = \frac{\Phi}{A \cdot \cos(\theta)}\).
04

Substitute Values and Calculate B

Substitute the known values into the rearranged equation: \(B = \frac{4.20 \times 10^{-4} \ \mathrm{Wb}}{8.96 \times 10^{-4} \ \mathrm{m^2} \cdot \frac{\sqrt{3}}{2}}\). Calculate \(B\) to find its magnitude.

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 Field
Magnetic fields are invisible forces that permeate space around a magnet or an electric current. They exert force on other magnets or charged particles within their vicinity. Imagine these fields as invisible lines flowing from one pole of a magnet to the other. This concept is essential in physics and engineering because it helps us understand how forces interact over a distance without direct contact. Magnetic fields have certain key properties:
  • Direction: By convention, the field direction is from north to south outside a magnet.
  • Magnitude: This reflects the strength of the field at any given point and is measured in teslas (T).
  • Influence: It affects currents, magnets, and other particles, aligning them with the field lines.
Understanding magnetic fields helps explain various phenomena in nature and technology, such as how compasses work or how electric motors function.
Rectangular Surface
A rectangular surface is a flat, two-dimensional shape with four sides and opposite sides of equal length. In problems involving magnetic fields, these surfaces often serve as the area through which magnetic flux is calculated. To determine the area of a rectangular surface, simply multiply the length by the width.
In our exercise, the dimensions of the rectangular surface are given: 2.80 cm by 3.20 cm. To work effectively with formulas, converting these units to square meters is crucial, as the International System of Units (SI) is the standard for scientific calculations. Here's a brief overview of surface area calculations:
  • Geometry: The rectangle鈥檚 straightforward shape makes area determination simpler.
  • Unit Conversion: Always convert to standard units (m虏) for consistency in calculations.
Once the area is found, it becomes a key factor in calculating magnetic flux.
Angle of Inclination
The angle of inclination is the angle between a reference direction and the direction of interest. In the context of our problem, the angle is crucial as it affects the calculation of magnetic flux through the surface.
The magnetic flux formula \(\Phi = B \cdot A \cdot \cos(\theta)\) \includes the cosine of the angle to account for how the magnetic field intersects with the surface.When dealing with angles:
  • The angle of 0掳 means the field is perpendicular to the surface, maximizing flux.
  • As the angle increases to 90掳, flux decreases since the field runs parallel to the surface.
  • In our exercise, a 30掳 angle requires calculating (\cos(30掳)), \which equals (\sqrt{3}/2).
Accurately determining this affects how we calculate the magnetic field strength required to achieve a certain flux.
Conversion of Units
Conversion of units is essential in physics to maintain consistency and ensure accurate calculations. In our exercise, converting units is necessary to find the area in the correct SI units and to simplify the magnetic flux calculations.
Here's how to handle unit conversion effectively:
  • Identify the units you start with and those you need to convert to, such as centimeters to meters.
  • Use the appropriate conversion factors: 1 cm = 0.01 m or 1 cm虏 = 10鈦烩伌 m虏.
  • Perform the conversion by multiplying or dividing, depending on the factor used.
Proper conversion ensures that all measurements are in compatible units, making calculations straightforward and reducing the risk of errors.

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

A deuteron (the nucleus of an isotope of hydrogen) has a mass of \(3.34 \times 10^{-27} \mathrm{kg}\) and a charge of \(+e .\) The deuteron travels in a circular path with a radius of 6.96 \(\mathrm{mm}\) in a magnetic field with magnitude 2.50 \(\mathrm{T}\) (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

An Electromagnetic Rail Gun. A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\vec { B }\) fills the region between the rails (Fig. \(P 27.74 )\) (a) Find the magnitude and direction of the net force on the con- ducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m ,\) find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\) . (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \(( 11.2 \mathrm { km } / \mathrm { s } ) .\) Let \(B = 0.80 \mathrm { T } , \quad I = 2.0 \times 10 ^ { 3 } \mathrm { A } , \quad m = 25 \mathrm { kg }\) and \(L = 50 \mathrm { cm } .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A proton \(\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)\) moves in a uniform magnetic field \(\vec{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{l}}\) . At \(t=0\) the proton has velocity components \(v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0\) and \(v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}\) (see Example 27.4\() .\) (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the \(+x\) -direction, \(\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}\) . (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t=T / 2\), where \(T\) is the period of the circular motion of the proton, what is the \(x\) -component of the displacement of the proton from its position at \(t=0 ?\)

A singly charged ion of \(^{7} \mathrm{Li}\) (an isotope of lithium) has a mass of \(1.16 \times 10^{-26} \mathrm{kg}\) . It is accelerated through a potential dif- ference of 220 \(\mathrm{V}\) and then enters a magnetic field with magnitude 0.723 T perpendicular to the path of the ion. What is the radius of the ion's path in the magnetic field?

An electron moves at \(2.50 \times 10^{6} \mathrm{m} / \mathrm{s}\) through a region in which there is a magnetic field of unspecified direction and magnitude \(7.40 \times 10^{-2} \mathrm{T}\) (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Magnetism

Read Explanation

Electrostatics

Read Explanation

Atoms and Radioactivity

Read Explanation

Electromagnetism

Read Explanation

Wave Optics

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.