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Chapter 27: Problem 11

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; \((\mathrm{b})\) at an angle of \(53.1^{\circ}\) from the \(+z\) -direction; \((\mathrm{c})\) in the \(+y\) -direction?

Short Answer

Expert verified
(a) 0.00306 T鈰卪虏, (b) 0.00184 T鈰卪虏, (c) 0 T鈰卪虏.

Step by step solution

01

Understand the Magnetic Flux Formula

Magnetic flux is calculated using the formula \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnetic field, \( A \) is the area of the circle, and \( \theta \) is the angle between the magnetic field and the normal (perpendicular) to the surface.
02

Calculate the Area of the Circle

The area \( A \) of the circle is given by \( A = \pi r^2 \). Substituting the radius \( r = 6.50 \, \text{cm} = 0.065 \, \text{m} \), we get:\[A = \pi (0.065)^2 = 0.0133 \, \text{m}^2\]
03

Step 3a: Calculate Magnetic Flux for Part (a)

For part (a), the magnetic field is in the \(+z\)-direction, which means \( \theta = 0^{\circ} \). Therefore, \( \cos(\theta) = \cos(0^{\circ}) = 1 \). The magnetic flux \( \Phi \) is:\[\Phi = 0.230 \, \text{T} \times 0.0133 \, \text{m}^2 \times 1 = 0.00306 \, \text{T} \cdot \text{m}^2\]
04

Step 3b: Calculate Magnetic Flux for Part (b)

For part (b), the angle \( \theta = 53.1^{\circ} \) from the \(+z\)-direction. So, \( \cos(53.1^{\circ}) = 0.6 \). The magnetic flux \( \Phi \) is:\[\Phi = 0.230 \, \text{T} \times 0.0133 \, \text{m}^2 \times 0.6 = 0.00184 \, \text{T} \cdot \text{m}^2\]
05

Step 3c: Calculate Magnetic Flux for Part (c)

For part (c), the magnetic field is in the \(+y\)-direction, which means the angle \( \theta = 90^{\circ} \) (perpendicular to the circle's normal in the \(z\)-axis). Thus, \( \cos(90^{\circ}) = 0 \).Therefore, the magnetic flux \( \Phi \) is:\[\Phi = 0.230 \, \text{T} \times 0.0133 \, \text{m}^2 \times 0 = 0 \, \text{T} \cdot \text{m}^2\]

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Key Concepts

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

Magnetic Field
The concept of a magnetic field is fundamental in understanding magnetic flux. A magnetic field is essentially a region around a magnetic material or a moving electric charge within which the force of magnetism acts. It is represented by the symbol
  • **B** and measured in Teslas (T).
  • In practical scenarios, the field could be due to a magnet, a current-carrying wire, or the Earth itself.
Magnetic fields have both magnitude and direction, making them a vector quantity. In exercises such as the one described, understanding the orientation of the magnetic field relative to a surface is key to determining the magnetic flux through that surface.
For instance, if the magnetic field is in the
  • **+z**-direction, it is perpendicular to the circular area lying in the **xy**-plane, leading to a straightforward multiplication in flux calculations.
  • On the other hand, if the field direction changes, the projection in the perpendicular component diminishes or vanishes.
Angle of Inclination
The angle of inclination, denoted
  • **胃**, represents the orientation of the magnetic field relative to the normal (perpendicular) to a surface.
  • In the context of calculating magnetic flux, this angle is critical because it determines how much of the magnetic field's strength actually "penetrates" through the given area.
  • The mathematical relationship is given by the cosine component **\( \cos(\theta) \)** in the magnetic flux formula.
The angle of inclination determines the effective exposure of the magnetic field to the area:
  • **0 degrees** inclination: The field is fully perpendicular to the surface and fully contributes to flux.
  • **90 degrees** inclination: The field is parallel to the surface with zero contribution to flux.
  • Any other angle: Part of the magnetic field contributes to the flux, calculated using **\( \cos(\theta) \)**.
Understanding this angle's role enhances the ability to evaluate different orientations of a magnetic setup.
Circular Area
The concept of area is equally crucial when determining magnetic flux. In this case, we focus on a circular area, which acts as the surface through which the magnetic field passes. The area of a
  • **circle** is given by the formula **\( A = \pi r^2 \)**, where **\( r \)** is the radius.
  • The value of the radius needs to be converted into meters if initially given in centimeters for the purpose of consistency with other SI units.
For example, a radius of 6.50 cm would be 0.065 m.
Calculating the area forms a foundation for determining magnetic flux because
  • **flux** is directly dependent on how much surface area the magnetic field is interacting with.
  • A larger area results in a greater opportunity for field lines to pass through, thereby increasing the flux given the same magnitude of the magnetic field.
Magnetic Field Direction
Direction is a key aspect of a magnetic field, describing which way the field lines point. The exercise presents different scenarios of the field's direction relative to a circular area.
  • When the magnetic field is aligned with the normal to a surface, as in the **+z**-direction scenario for **Part (a)**, the field is maximally effective for calculating flux.
  • In contrast, when the field is parallel to the surface, like in **Part (c)** in the **+y**-direction, no field lines pass through the surface, resulting in zero flux.
Moreover, components of the magnetic field along different directions can lead to varying angles of interaction, which are covered in the angle of inclination section. Thus, understanding the directional aspect helps with comprehending the significance of orientation in influencing the net magnetic effect experienced by a surface.

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Most popular questions from this chapter

A particle with charge \(q\) is moving with speed \(v\) in the \(- y\) -direction. It is moving in a uniform magnetic field \(\vec { B } =\) \(B _ { x } \hat { i } + B _ { y } \hat { J } + B _ { z } \hat { k } .\) (a) What are the components of the force \(\vec { \boldsymbol { F } }\) exerted on the particle by the magnetic field? (b) If \(q > 0 ,\) what must the signs of the components of \(\vec { B }\) be if the components of \(\vec { \boldsymbol { F } }\) are all nonnegative? (c) If \(q < 0\) and \(B _ { x } = B _ { y } = B _ { z } > 0 ,\) find the direction of \(\vec { \boldsymbol { F } }\) and find the magnitude of \(\vec { \boldsymbol { F } }\) in terms of \(| q | , v ,\) and \(B _ { x }\)

Paleoclimate. Climatologists can determine the past temperature of the earth by comparing the ratio of the isotope oxygen-18 to the isotope oxygen-16 in air trapped in ancient ice sheets, such as those in Greenland. In one method for separating these isotopes, a sample containing both of them is first singly ionized (one electron is removed) and then accelerated from rest through a potential difference \(V\) . This beam then enters a magnetic field \(B\) at right angles to the field and is bent into a quarter-circle. A particle detector at the end of the path measures the amount of each isotope. (a) Show that the separation \(\Delta r\) of the two isotopes at the detector is given by $$\Delta r = \frac { \sqrt { 2 e V } } { e B } \left( \sqrt { m _ { 18 } } - \sqrt { m _ { 16 } } \right)$$ where \(m _ { 16 }\) and \(m _ { 18 }\) are the masses of the two oxygen isotopes, (b) The measured masses of the two isotopes are \(2.66 \times\) \(10 ^ { - 26 } \mathrm { kg } \left( ^ { 16 } \mathrm { O } \right)\) and \(2.99 \times 10 ^ { - 26 } \mathrm { kg } ( 8 \mathrm { O } ) .\) If the magnetic field is \(0.050 \mathrm { T } ,\) what must be the accelerating potential \(V\) so that these two isotopes will be separated by 4.00\(\mathrm { cm }\) at the detector?

A particle of charge \(q > 0\) is moving at speed \(v\) in the \(+ z\) -direction through a region of uniform magnetic field \(\vec { \boldsymbol { B } }\) . The magnetic force on the particle is \(\vec { \boldsymbol { F } } = F _ { 0 } ( 3 \hat { \boldsymbol { \imath } } + 4 \hat { \boldsymbol { J } } ) ,\) where \(F _ { 0 }\) is a positive constant. (a) Determine the components \(B _ { x } , B _ { y } ,\) and \(B _ { z }\) , or at least as many of the three components as is possible from the information given. (b) If it is given in addition that the magnetic field has magnitude \(6 F _ { 0 } / q v ,\) determine as much as you can about the remaining components of \(\vec { B } .\)

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\vec{\boldsymbol{B}}=-(1.25 \mathrm{T}) \hat{\boldsymbol{k}}\) . The magnetic force on the particle is measured to be \(\vec{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{N}\right) \hat{\imath}+(7.40 \times\) \(10^{-7} \mathrm{N} ) \hat{J}\) . (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v} \cdot\) What is the angle between \(\vec{v}\) and \(\vec{F} ?\)

A magnetic field exerts a torque \(\tau\) on a round current carrying loop of wire. What will be the torque on this loop (in terms of \(\tau\) if its diameter is tripled?
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