/*! 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 circular area with a radius of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A circular area with a radius of 6.50 \(\mathrm{cm}\) lies in the \(x\) -y plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field \(B=0.230 \mathrm{T}\) that points (a) in the \(+z\) direction? (b) at an angle of \(53.1^{\circ}\) from the \(+z\) direction? (c) in the \(+y\) direction?

Short Answer

Expert verified
(a) 0.00306 Wb, (b) 0.00184 Wb, (c) 0 Wb.

Step by step solution

01

Calculate Area of the Circle

First, calculate the area of the circular area. The formula for the area of a circle is \(A = \pi r^2\), where \(r\) is the radius of the circle. Here, \(r = 6.50 \, \mathrm{cm} = 0.065 \, \mathrm{m}\). Substituting, we have:\[A = \pi (0.065)^2 \approx 0.0133 \, \mathrm{m^2}\]
02

Calculate Magnetic Flux for (a) - Field in +z Direction

The magnetic flux \(\Phi\) for a magnetic field \(B\) perpendicular to the surface (along the normal) is given by \(\Phi = B \cdot A\). Since the magnetic field is in the \(+z\) direction and the circle is in the \(x-y\) plane, the angle \(\theta\) between them is \(0^\circ\). Thus, \[ \Phi = B \cdot A \cdot \cos(0^\circ) = 0.230 \, \mathrm{T} \times 0.0133 \, \mathrm{m^2} \approx 0.00306 \, \mathrm{Wb}\]
03

Calculate Magnetic Flux for (b) - Field at 53.1° from +z Direction

In this case, the angle \(\theta\) between the magnetic field and the normal to the circle is \(53.1^\circ\). Thus, the magnetic flux is calculated as:\[ \Phi = B \cdot A \cdot \cos(53.1^\circ) = 0.230 \, \mathrm{T} \times 0.0133 \, \mathrm{m^2} \times \cos(53.1^\circ) \approx 0.00184 \, \mathrm{Wb}\]
04

Calculate Magnetic Flux for (c) - Field in +y Direction

When the magnetic field is in the \(+y\) direction, it is perpendicular to the normal of the circular area, as the normal is in the \(+z\) direction. Therefore, the angle \(\theta = 90^\circ\). The magnetic flux is:\[ \Phi = B \cdot A \cdot \cos(90^\circ) = 0.230 \, \mathrm{T} \times 0.0133 \, \mathrm{m^2} \times 0 = 0 \, \mathrm{Wb} \]

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.

Magnetic Field
The concept of the magnetic field is central in understanding how forces affect charged particles in motion. A magnetic field represents a region where a magnetic force can be detected. It is often generated by moving electric charges or magnetic dipoles and can be uniform or vary in strength and direction.
Magnetic field lines provide a way to visualize a magnetic field. They start from the north pole and end at the south pole of a magnet. The direction of these lines indicates the direction of the magnetic force a north pole of a magnet would experience. The density of these lines reflects the strength of the magnetic field: the closer the lines, the stronger the field.
  • The magnetic field can be measured in units called Tesla (T).
  • A uniform magnetic field means the magnetic lines are parallel and equally spaced.
In the context of calculating magnetic flux through a circle, the field could have different orientations relative to the circle, affecting the amount of flux.
Circular Area
A circular area is simply the two-dimensional space enclosed by a circle. The size of this area plays an important role when calculating magnetic flux. The larger the area, the more flux it can capture, assuming other things like the magnetic field strength and orientation remain constant.
The formula for determining the area of a circle is given by:\[A = \pi r^2\]where \(r\) is the radius. If it's essential to work in consistent units, remember to convert all measurements to meters or the standard unit you're using.
  • Area is measured in square meters \( \mathrm{m^2} \) in physics.
  • A precise measurement is crucial for accurate calculations.
By knowing the circle's area, you can calculate how much of a magnetic field passes through it, when combined with the properties of the magnetic field and its angle of inclination.
Angle of Inclination
The angle of inclination, expressed as \( \theta \), is the angle between the magnetic field line and the normal (perpendicular) to the surface of the circular area. This angle is crucial because it affects the magnitude of the magnetic flux.
Magnetic flux \( \Phi \) through a surface is calculated by the formula:\[\Phi = B \cdot A \cdot \cos(\theta)\]where \(B\) is the magnetic field strength, \(A\) is the area, and \(\theta\) is this angle of inclination.
  • When the field is perpendicular to the surface (\(\theta = 0^\circ \)), \(\cos(0^\circ) = 1\), indicating maximum flux.
  • When parallel (\(\theta = 90^\circ\)), \(\cos(90^\circ) = 0\), meaning no flux passes through.
The inclination angle tells us how much of the magnetic field actually penetrates the surface, effectively tuning the amount of flux through geometric orientation.

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

Off to Europe! You plan to take your hair blower to Europe, where the electrical outlets put out 240 \(\mathrm{V}\) instead of the 120 \(\mathrm{V}\) seen in the United States. The blower puts out 1600 \(\mathrm{W}\) at 120 \(\mathrm{V}\) . (a) What could you do to operate your blower via the 240 \(\mathrm{V}\) line in Europe? (b) What current will your blower draw from a European outlet? (c) What resistance will your blower appear to have when operated at 240 \(\mathrm{V} ?\)

You're driving at 95 \(\mathrm{km} / \mathrm{h}\) in a direction \(35^{\circ}\) east of north, in a region where the earth's magnetic field of \(5.5 \times 10^{-5} \mathrm{T}\) is horizontal and points due north. If your car measures 1.5 \(\mathrm{m}\) from its underbody to its roof, calculate the induced emf between roof and underbody. (You can assume the sides of the car are straight and vertical.) Is the roof of the car at a higher or lower potential than the underbody?

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12 \(\mathrm{cm}^{2}\) is rotated from a position where its plane is perpendicular to the earth's magnetic field to one where its plane is parallel to the field. The rotation takes 0.040 s. The earth's magnetic field at the location of the laboratory is \(6.0 \times 10^{-5} \mathrm{T.}\) (a) What is the total magnetic flux through the coil before it is rotated? After it is rotated? (b) What is the average emf induced in the coil?

A 5.00\(\mu \mathrm{F}\) capacitor is initially charged to a potential of 16.0 \(\mathrm{V}\) . It is then connected in series with a 3.75 \(\mathrm{mH}\) inductor. (a) What is the total energy stored in this circuit? (b) What is the maximum current in the inductor? What is the charge on the capacitor plates at the instant the current in the inductor is maximal?

A 1.50 \(\mathrm{mH}\) inductor is connected in series with a dc battery of negligible internal resistance, a 0.750 \(\mathrm{k} \Omega\) resistor, and an open switch. How long after the switch is closed will it take for (a) the current in the circuit to reach half of its maximum value, (b) the energy stored in the inductor to reach half of its maximum value? (Hint: You will have to solve an exponential equation.)

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.