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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; \((b)\) 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 Magnetic Flux

Magnetic flux (\(\Phi\) ) through a surface is given by the formula \(\Phi = B \cdot A \cdot \cos(\theta)\), where \(B\) is the magnetic field, \(A\) is the area, and \(\theta\) is the angle between the magnetic field and the normal (perpendicular) to the surface.
02

Calculate the Area of the Circle

Given the radius \(r = 6.50\, \mathrm{cm}\), convert it to meters: \(r = 0.0650\, \mathrm{m}\). The area \(A\) of a circle is \(A = \pi r^2\). Substituting the radius, we get \(A = \pi \, (0.0650)^2 = 0.0133 \, \mathrm{m^2}\).
03

Calculate Magnetic Flux for Case (a)

For case (a), the magnetic field is perpendicular to the plane of the circle (\(\theta = 0^{\circ}\)). Thus, \(\cos(0^{\circ}) = 1\). Substituting the values into the flux formula \(\Phi = 0.230 \times 0.0133 \times 1 = 0.00306 \, \mathrm{T\cdot m^2}\).
04

Calculate Magnetic Flux for Case (b)

For case (b), the angle \(\theta = 53.1^{\circ}\). Calculate \(\cos(53.1^{\circ})\) which is approximately \(0.6018\). Substituting into the formula \(\Phi = 0.230 \times 0.0133 \times 0.6018 = 0.00184 \, \mathrm{T\cdot m^2}\).
05

Calculate Magnetic Flux for Case (c)

For case (c), the magnetic field is parallel to the plane of the circle (\(\theta = 90^{\circ}\)). Thus, \(\cos(90^{\circ}) = 0\). Therefore, the magnetic flux is \(\Phi = 0.230 \times 0.0133 \times 0 = 0 \, \mathrm{T\cdot m^2}\).

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Key Concepts

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

Magnetic Field
The magnetic field, often denoted by \( B \), is a vector field surrounding magnets and electric currents, and it exerts magnetic force on moving charges. The strength of a magnetic field is measured in teslas (\( \text{T} \)). It can have different orientations, and its relationship with a surface or object is crucial in determining magnetic effects.
  • If the magnetic field is perpendicular to a plane, it maximizes the interaction causing the highest possible magnetic flux for a given magnitude of \( B \).
  • If the magnetic field is parallel to the plane, the effect is minimized, as seen in cases like a magnetic field in the \( +y \)-direction with a surface lying on the \( xy \)-plane.
  • Any intermediate angle between the field and the surface needs to be considered through its cosine function value, as used in the calculation of magnetic flux through non-perpendicular angles.
Understanding the orientation and interactions of magnetic fields helps in various applications from electrical engineering to geophysics.
Area of a Circle
The area of a circle is a fundamental concept in geometry, critical to calculating quantities like magnetic flux through circular surfaces. It is calculated using the formula: \[ A = \pi r^2 \] where \( r \) is the radius of the circle. This formula implies that the area grows with the square of the radius.
  • For a radius of 6.50 cm, converting to meters is necessary for consistency with standard units in physics, giving us \( r = 0.0650 \ \text{m} \).
  • Plugging this into the formula yields an area of approximately \( A = 0.0133 \ \mathrm{m^2} \).
When calculating the magnetic flux, knowing the accurate area is critical, since any error can lead to misinterpretation of magnetic interactions.
Angle in Physics
In physics, angles play a crucial role in understanding how forces and fields interact with surfaces. The angle \( \theta \) is particularly important in calculations of magnetic flux, where it defines the orientation between a magnetic field and the normal to a surface.
  • When the angle is \( 0^{\circ} \), the magnetic field is perpendicular to the plane, leading to maximum flux through the surface since \( \cos(0^{\circ}) = 1 \).
  • At an angle of \( 90^{\circ} \), the field is parallel to the surface, and the magnetic flux equals zero because \( \cos(90^{\circ}) = 0 \).
  • Intermediate angles require the cosine of the angle, such as \( \cos(53.1^{\circ}) \approx 0.6018 \) in our example, to calculate the effective contribution of the magnetic field to the flux.
The cosine function captures how much of the field is "effective" in generating flux over the area of interest. Thus, angle measurement and calculations ensure that magnetic interaction assessments are precise.

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Most popular questions from this chapter

Magnetic Moment of the Hydrogen Atom. In the Bohr model of the hydrogen atom (see Section \(38.5 ),\) in the lowest energy state the electron orbits the proton at a speed of \(2.2 \times\) \(10^{6} \mathrm{m} / \mathrm{s}\) in a circular orbit of radius \(5.3 \times 10^{-11} \mathrm{m}\) (a) What is the orbital period of the electron? (b) If the orbiting electron is considered to be a current loop, what is the current \(I ?\) (c) What is the magnetic moment of the atom due to the motion of the electron?

An insulated wire with mass \(m=5.40 \times 10^{-5} \mathrm{kg}\) is bent into the shape of an inverted U such that the horizontal part has a length \(l=15.0 \mathrm{cm} .\) The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 \(\mathrm{cm}\) of each end below the mercury's surface. The entire structure is in a region containing a uniform \(0.00650-\mathrm{T}\) magnetic field directed into the page (Fig. 27.71\()\) . An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a \(1.50-\mathrm{V}\) battery and a switch \(\mathrm{S}\) . When switch \(\mathrm{S}\) is closed, the wire jumps 35.0 \(\mathrm{cm}\) into the air, measured from its initial position. (a) Determine the speed \(v\) of the wire as it leaves the mercury. (b) Assuming that the current \(I\) through the wire was constant from the time the switch was closed until the wire left the mercury, determine \(I\) (c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 \(\mathrm{km} / \mathrm{s}\) im the \(+x\) -direction experiences a force of \(2.25 \times 10^{-16} \mathrm{N}\) in the \(+y\) -direction, and an electron moving at 4.75 \(\mathrm{km} / \mathrm{s}\) in the \(-\mathrm{z}\) -direction experiences a force of \(8.50 \times 10^{-16} \mathrm{N}\) . (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\) -direction at 3.2 \(\mathrm{km} / \mathrm{s} ?\)

A particle of charge \(q>0\) is moving at speed \(v\) in the \(+z\) -direction through a region of uniform magnetic field \(\overrightarrow{\boldsymbol{B}} .\) The magnetic force on the particle is \(\overrightarrow{\boldsymbol{F}}=F_{0}(3 \hat{\imath}+4 \hat{\jmath}),\) where \(\boldsymbol{F}_{0}\) 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 \(\overrightarrow{\boldsymbol{B}}\) .

A beam of protons traveling at 1.20 \(\mathrm{km} / \mathrm{s}\) enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular- to its original direction (Fig. 27.48\()\) . The beam travels a distance of 1.18 \(\mathrm{cm}\) while in the field. What is the magnitude of the magnetic field?
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