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A magnetic field has a magnitude of \(0.078 \mathrm{T}\) and is uniform over a circular surface whose radius is \(0.10 \mathrm{m}\). The field is oriented at an angle of \(\phi=25^{\circ}\) with respect to the normal to the surface. What is the magnetic flux through the surface?

Short Answer

Expert verified
The magnetic flux through the surface is approximately 0.00222 Wb.

Step by step solution

01

Understand Magnetic Flux Formula

Magnetic flux \( \Phi_B \) through a surface is defined as \( \Phi_B = B \cdot A \cdot \cos(\phi) \), where \( B \) is the magnetic field strength, \( A \) is the area of the surface, and \( \phi \) is the angle between the magnetic field and the normal to the surface.
02

Calculate Area of the Circular Surface

The area \( A \) of a circle is given by \( A = \pi R^2 \), where \( R = 0.10 \mathrm{m} \) is the radius of the circle. Thus, \( A = \pi (0.10)^2 = 0.0314 \mathrm{m}^2 \).
03

Identify Given Values

We have the magnetic field \( B = 0.078 \mathrm{T} \), the angle \( \phi = 25^{\circ} \), and the calculated area \( A = 0.0314 \mathrm{m}^2 \).
04

Substitute Values into Magnetic Flux Formula

Substitute the known values into the magnetic flux formula: \( \Phi_B = 0.078 \mathrm{T} \times 0.0314 \mathrm{m}^2 \times \cos(25^{\circ}) \).
05

Calculate Cosine of the Angle

Calculate \( \cos(25^{\circ}) \) using a calculator, which gives approximately \( 0.9063 \).
06

Compute Magnetic Flux

Now calculate \( \Phi_B = 0.078 \times 0.0314 \times 0.9063 \approx 0.00222 \mathrm{Wb} \).

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Key Concepts

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

Magnetic Field Strength
The magnetic field strength, often denoted as \( B \), is a measure of the magnitude of a magnetic field at a given point. It is usually measured in Tesla (T). Think of it as a force field surrounding a magnet or a current-carrying wire that can exert forces on other magnets or moving electric charges within the field. Understanding this measure helps us predict how the magnetic field will interact with surrounding objects.
In the given exercise, the magnetic field strength is provided as \( 0.078 \text{ T} \), which indicates a relatively moderate strength field compared to everyday sources like household magnets or the Earth's magnetic field. Ensuring clarity on this concept helps to determine how significantly the field will influence the magnetic flux. Without a clear grasp of magnetic field strength, it can become challenging to evaluate how different setups, such as a different orientation or field medium, may alter the results in practical applications.
Remember:
  • In the magnetic flux formula \( \Phi_B = B \cdot A \cdot \cos(\phi) \), \( B \) directly influences the total magnetic flux.
  • A higher \( B \) means a more robust force acting across a surface area, resulting in a greater calculated flux value.
Circular Area Calculation
Understanding how to calculate the area of a circular surface is crucial for applying many physics concepts, including magnetic flux. The area \( A \) of a circle is determined using the formula \( A = \pi R^2 \), where \( R \) is the circle's radius. In our exercise, the radius of the circular surface is \( 0.10 \text{ m} \), leading to an area calculation of \( A = \pi (0.10)^2 = 0.0314 \text{ m}^2 \).
This calculation is fundamental because the area directly impacts the magnitude of the magnetic flux. A larger area would correspondingly capture more of the magnetic field, leading to a higher flux.
When calculating circle areas, remember:
  • Ensure the radius is correctly measured in meters, as this conforms to the standard units used in scientific calculations.
  • Always use \( \pi \approx 3.1416 \) for accurate area computation.
It's also worth noting that any error in the measured radius or computation can lead to significant errors in subsequent flux calculations, making precision paramount.
Angle of Orientation
The angle of orientation \( \phi \) in this context relates to how the circular surface is positioned relative to the magnetic field lines. Specifically, it's the angle between the magnetic field direction and the normal (perpendicular) to the surface of interest. In our exercise, \( \phi = 25^{\circ} \).
Understanding this angle is vital because it determines how much of the magnetic field effectively "penetrates" the surface, influencing the magnetic flux. The smaller the angle between the field lines and the normal to the surface, the greater the flux through the surface. This is mathematically represented by the cosine term in the flux formula \( \Phi_B = B \cdot A \cdot \cos(\phi) \).
Here’s why the angle is essential:
  • \( \cos(\phi) \) adjusts the effective surface area interacting with the magnetic field.
  • An angle of \( 0^{\circ} \) would result in the maximum possible flux (since \( \cos(0^{\circ}) = 1 \)).
  • As the angle increases towards \( 90^{\circ} \), \( \cos(\phi) \) decreases to zero, reducing the flux to a minimum.
In practical scenarios, adjusting the angle of orientation can be an effective way to control the amount of magnetic flux that a surface encounters, which is an important consideration in designing electromagnetic devices and systems.

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Most popular questions from this chapter

A standard door into a house rotates about a vertical axis through one side, as defined by the door's hinges. A uniform magnetic field is parallel to the ground and perpendicular to this axis. Through what angle must the door rotate so that the magnetic flux that passes through it decreases from its maximum value to one-third of its maximum value?

A \(5.40 \times 10^{-5} \mathrm{H}\) solenoid is constructed by wrapping 65 turns of wire around a cylinder with a cross-sectional area of \(9.0 \times 10^{-4} \mathrm{m}^{2} .\) When the solenoid is shortened by squeezing the turns closer together, the inductance increases to \(8.60 \times 10^{-5}\) H. Determine the change in the length of the solenoid.

A generator uses a coil that has 100 turns and a \(0.50-\) T magnetic field. The frequency of this generator is \(60.0 \mathrm{Hz}\), and its emf has an rms value of \(120 \mathrm{V}\). Assuming that each turn of the coil is a square (an approximation), determine the length of the wire from which the coil is made.

You need to design a \(60.0-\mathrm{Hz}\) ac generator that has a maximum emf of \(5500 \mathrm{V}\). The generator is to contain a 150 -turn coil that has an area per turn of \(0.85 \mathrm{m}^{2} .\) What should be the magnitude of the magnetic field in which the coil rotates?

A loop of wire has the shape shown in the drawing. The top part of the wire is bent into a semicircle of radius \(r=0.20 \mathrm{m} .\) The normal to the plane of the loop is parallel to a constant magnetic field \(\left(\phi=0^{\circ}\right)\) of magnitude 0.75 T. What is the change \(\Delta \Phi\) in the magnetic flux that passes through the loop when, starting with the position shown in the drawing, the semicircle is rotated through half a revolution?

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