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Chapter 27: Problem 9

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 \(B=0.230 \mathrm{~T}\) (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) For the field in the +z direction, the magnetic flux is \(0.0866 \, \mathrm{Wb}\). (b) For the field at an angle of \(53.1^\circ\) from the +z direction, the magnetic flux is \(0.0514 \, \mathrm{Wb}\). (c) For the field in the +y direction, the magnetic flux is 0 as the angle between the field and area vector is \(90^\circ\).

Step by step solution

01

Understanding and Calculating the Area

Firstly, calculate the area \(A\) of the circular area using the formula \(A = \pi r^2\), where \( r = 6.50 \, \mathrm{cm} = 0.065 \, \mathrm{m}\). This will be used for all parts of the problem.
02

Calculating the Magnetic Flux for a Field in the +z Direction

To calculate the magnetic flux \(\Phi\) through the circle due to a uniform magnetic field \(B\) in the \(+z\) direction, use the formula \(\Phi = BA\cos\theta\), with \(\theta = 0^\circ\) as the field is in the +z direction and angle between the area vector (outward to +z direction) and magnetic field is 0 degrees. Substitute the values for \(B\), \(A\) and \(\cos\theta\).
03

Calculating the Magnetic Flux for a Field at an Angle from the +z Direction

To calculate \(\Phi\) due to \(B\) at an angle of \(53.1^\circ\) from the \(+z\) direction, still use the formula \(\Phi = BA\cos\theta\) but now \(\theta = 53.1^\circ\). Substitute the values for \(B\), \(A\) and \(\cos\theta\).
04

Calculating the Magnetic Flux for a Field in the +y Direction

To calculate \(\Phi\) due to \(B\) in the \(+y\) direction, still use the formula \(\Phi = BA\cos\theta\) but now \(\theta = 90^\circ\) as the field is perpendicular to area vector. Substitute the values for \(B\), \(A\) and \(\cos\theta\).

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Key Concepts

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

Circular Area
When we talk about a circular area in the context of magnetic flux, we're referring to the two-dimensional space that is bounded by the circumference of a circle. The "area" is critical in calculating the magnetic flux because it represents the surface through which the magnetic field passes. In this problem, our circle has a radius of 6.50 cm. In order to calculate the area of this circle, we use the formula for the area of a circle, which is:
  • \( A = \pi r^2 \)
  • where \( r = 6.50 \text{ cm} = 0.065 \text{ m} \)
Substituting the radius into the formula gives us the area in square meters. This area is constant across all parts of the exercise because the shape and size of the circular region do not change.
Uniform Magnetic Field
A uniform magnetic field means that the strength and direction of the magnetic field are consistent across the entire region. In our exercises, the magnetic field has been given as 0.230 T (Tesla), which represents a strong and constant magnetic influence. Uniformity simplifies calculations because it ensures the magnetic flux through different areas does not vary due to field strength changes.In formulaic terms, the magnetic flux \( \Phi \) is calculated using:
  • \( \Phi = B \cdot A \cdot \cos(\theta) \)
  • \( B \) represents the magnetic field's magnitude, which is the same across the entire area.
This simplification assumes no distortion or alteration of the field as it passes through different sections of the circular area. Adjustments to calculations are needed only for changes in angle \( \theta \) but not for the field's intrinsic characteristics.
Angle of Inclination
The angle of inclination refers to the angle \( \theta \) between the magnetic field direction and the perpendicular (normal) vector to the surface of the circular area. This angle affects how much of the magnetic field "goes through" the surface.Here, understanding the angle allows us to determine the effective component of the magnetic field that passes through the circle:
  • When \( \theta = 0掳 \), the magnetic field is fully aligned with the normal, maximizing flux because \( \cos(0掳) = 1 \).
  • For \( \theta = 53.1掳 \), only part of the field contributes to the flux, since \( \cos(53.1掳) \) is less than 1.
  • With \( \theta = 90掳 \), the field is perpendicular to the normal, leading to no net flux contribution, as \( \cos(90掳) = 0 \).
The formula \( \Phi = BA \cos(\theta) \) showcases how crucial the angle is, as it can substantially change the outcome of the flux calculations.
Magnetic Field Direction
The direction of the magnetic field is crucial as it defines its interaction with the circular area. Different directions can lead to different values of the magnetic flux as the effective field component changes. The exercise explores three specific directions:
  • The \(+z\) direction: Here, the field is parallel to the normal of the circle surface, making flux calculations straightforward with maximum value.
  • From \(+z\) at an angle: The field is partially aligned, reducing the effective flux.
  • The \(+y\) direction: The field is perpendicular to the normal, causing zero flux.
Knowing the direction helps us employ \( \cos(\theta) \) correctly in flux computations, calibrating it according to the precise alignment of the field against the circular area's normal vector. Different angles illustrate varying physical scenarios that manifest through changes in calculation outcomes.

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Most popular questions from this chapter

The plane of a \(5.0 \mathrm{~cm} \times 8.0 \mathrm{~cm}\) rectangular loop of wire is parallel to a 0.19 T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

The large magnetic fields used in MRI can produce forces on electric currents within the human body. This effect has been proposed as a possible method for imaging "biocurrents" flowing in the body, such as the current that flows in individual nerves. For a magnetic field strength of \(2 \mathrm{~T}\), estimate the magnitude of the maximum force on a 1-mm-long segment of a single cylindrical nerve that has a diameter of \(1.5 \mathrm{~mm} .\) Assume that the entire nerve carries a current due to an applied voltage of \(100 \mathrm{mV}\) (that of a typical action potential). The resistivity of the nerve is \(0.6 \Omega \cdot \mathrm{m}\). (a) \(6 \times 10^{-7} \mathrm{~N} ;\) (b) \(1 \times 10^{-6} \mathrm{~N} ;\) (c) \(3 \times 10^{-4} \mathrm{~N}\) (d) \(0.3 \mathrm{~N}\).

Suppose the electric field between the plates in Fig. 27.22 is \(1.88 \times 10^{4} \mathrm{~V} / \mathrm{m}\) and the magnetic field in both regions is \(0.682 \mathrm{~T}\). If the source contains the three isotopes of krypton, \({ }^{82} \mathrm{Kr},{ }^{84} \mathrm{Kr},\) and \({ }^{86} \mathrm{Kr},\) and the ions are singly charged, find the distance between the lines formed by the three isotopes on the particle detector. Assume the atomic masses of the isotopes (in atomic mass units) are equal to their mass numbers, \(82,84,\) and \(86 .\) (One atomic mass unit \(=1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{~kg} .\) )

In a 1.25 T magnetic field directed vertically upward, a particle having a charge of magnitude \(8.50 \mu \mathrm{C}\) and initially moving northward at \(4.75 \mathrm{~km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.

A \(150 \mathrm{~g}\) ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125 \mathrm{~m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude \(0.250 \mathrm{~T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.
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