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Chapter 21: Problem 1

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) Flux is about 0.00306 T路m虏, (b) Flux is about 0.00184 T路m虏, (c) Flux is 0 T路m虏.

Step by step solution

01

Understanding Magnetic Flux

Magnetic flux through a surface is given by the formula \( \Phi = B \cdot A \cdot \cos \theta \), where \(B\) is the magnetic field strength, \(A\) is the area of the surface, and \(\theta\) is the angle between the magnetic field lines and the normal (perpendicular) to the surface. We will use this formula to find the flux in each case.
02

Calculate Area of the Circle

First, we calculate the area of the circle using the formula for the area of a circle, \( A = \pi r^2 \). Given radius \( r = 6.50 \text{ cm} = 0.065 \text{ m} \), we have:\[ A = \pi (0.065)^2 \approx 0.0133 \text{ m}^2 \].
03

Case (a): Magnetic Field in +z Direction

Here, the magnetic field is perpendicular to the plane of the circle, so \( \theta = 0 \degree \). Therefore, the cosine term \( \cos \theta = \cos 0 = 1 \). Thus, the magnetic flux is:\[ \Phi = B \cdot A \cdot \cos 0 = 0.230 \times 0.0133 \times 1 \approx 0.00306 \text{ T} \cdot \text{m}^2. \]
04

Case (b): Magnetic Field at 53.1掳 from +z Direction

In this case, \( \theta = 53.1 \degree \). The magnetic flux is calculated by:\[ \Phi = B \cdot A \cdot \cos 53.1 \approx 0.230 \times 0.0133 \times \cos 53.1 \approx 0.230 \times 0.0133 \times 0.6 \approx 0.00184 \text{ T} \cdot \text{m}^2. \]
05

Case (c): Magnetic Field in +y Direction

Here, the magnetic field is parallel to the plane of the circle, so \( \theta = 90 \degree \). Therefore, \( \cos 90 = 0 \) and the magnetic flux is:\[ \Phi = B \cdot A \cdot \cos 90 = 0.230 \times 0.0133 \times 0 = 0 \text{ T} \cdot \text{m}^2. \]

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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 central to understanding magnetic phenomena. A magnetic field originates from magnets or electric currents and exerts a force on nearby ferromagnetic materials or moving electric charges. The strength of a magnetic field is measured in teslas (T), where higher values indicate a stronger force exerted by the field. Magnetic fields have both magnitude and direction, making them vector quantities, which is crucial when calculating interactions like magnetic flux. Understanding the directionality is especially important, as it influences how we calculate the angle of incidence between a magnetic field and a given surface.
Angle of Incidence
The angle of incidence in magnetic flux calculations refers to the angle between the magnetic field lines and a line normal to the surface of interest, which in this case is the circular area. This angle is denoted by \( \theta \) in the magnetic flux equation \( \Phi = B \cdot A \cdot \cos \theta \). The value of \( \theta \) determines how much of the magnetic field actually passes perpendicularly through the surface.
The cosine component \( \cos \theta \) scales the effect of the magnetic field based on its alignment with the surface's normal.
  • If \( \theta = 0 \degree \) (field perpendicular to the surface), \( \cos \theta = 1 \).
  • If \( \theta = 90 \degree \) (field parallel to the surface), \( \cos \theta = 0 \).
This parameter is essential to understanding how orientation affects magnetic flux.
Circular Area
A circular area in physics is often involved in calculations concerning curves and symmetry, especially when examining how fields interact with surfaces. The formula to calculate the area \( A \) of a circle is \( A = \pi r^2 \), where \( r \) is the radius. In our magnetic flux exercise, we've considered a circle with a radius of \( 6.50 \text{ cm} \) which works out to an area of about \( 0.0133 \text{ m}^2 \). This size allows for straightforward flux calculations when placed in different orientations relative to a magnetic field.
The circle鈥檚 placement and rotation relative to a magnetic field鈥檚 direction fundamentally alter the magnetic flux calculation. Adjusting the orientation changes the angle of incidence, thereby altering the effective `catching` area of the field.
Physics Problem Solving
Physics problem-solving often involves a methodical approach. Let's outline a strategic process to solve issues like this magnetic flux problem:
  • Begin by breaking down the problem: What are we asked to find? Here, it's the magnetic flux through a circular area.
  • Identify the known elements: We have the magnetic field strength \( B = 0.230 \text{ T} \) and the radius of the circle \( r = 6.50 \text{ cm} \) to derive its area.
  • Understand the context: Recognize the orientation to determine \( \theta \) and calculate \( \cos \theta \).
  • Apply relevant formulas: Plug these into the magnetic flux equation to get the solution.
This structured tactic allows you to connect the dots from abstract concepts to practical calculations, ensuring you reliably reach the right solution.

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Most popular questions from this chapter

A transformer consists of 275 primary windings and 834 secondary windings. If the potential difference across the primary coil is \(25.0 \mathrm{~V}\) (a) what is the voltage across the secondary coil, and (b) what is the current in the secondary coil if it is connected across a \(125 \Omega\) resistor?

A step-up transformer. A transformer connected to a \(120 \mathrm{~V}\) (rms) ac line is to supply \(13,000 \mathrm{~V}\) (rms) for a neon sign. To reduce the shock hazard, a fuse is to be inserted in the primary circuit and is to blow when the rms current in the secondary circuit exceeds \(8.50 \mathrm{~mA}\). (a) What is the ratio of secondary to primary turns of the transformer? (b) What power must be supplied to the transformer when the rms secondary current is \(8.50 \mathrm{~mA} ?\) (c) What current rating should the fuse in the primary circuit have?

When the current in a toroidal solenoid is changing at a rate of \(0.0260 \mathrm{~A} / \mathrm{s}\), the magnitude of the induced emf is \(12.6 \mathrm{mV}\). When the current equals \(1.40 \mathrm{~A}\), the average flux through each turn of the solenoid is 0.00285 Wb. How many turns does the solenoid have?

BIO Measuring blood flow. Blood contains positive and negative ions and so is a conductor. A blood vessel, therefore, can be viewed as an electrical wire. We can even picture the flowing blood as a series of parallel conducting slabs whose thickness is the diameter \(d\) of the vessel moving with speed \(v\). (See Figure \(21.59 .\) ) (a) If the blood vessel is placed in a magnetic field \(\overrightarrow{\boldsymbol{B}}\) perpendicular to the vessel, as in the figure, show that the motional potential difference induced across it is \(\mathcal{E}=v B d\). (b) If you expect that the blood will be flowing at \(15 \mathrm{~cm} / \mathrm{s}\) for a vessel \(5.0 \mathrm{~mm}\) in diameter, what strength of magnetic field will you need to produce a potential difference of \(1.0 \mathrm{mV} ?\) (c) Show that the volume rate of flow \((R)\) of the blood is equal to \(R=\pi \mathcal{E} d / 4 B\). (Note: Although the method developed here is useful in measuring the rate of blood flow in a vessel, it is of limited use in surgery because measurement of the potential \(\mathcal{E}\) must be made directly across the vessel.)

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?
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