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Chapter 27: Problem 12

A horizontal rectangular surface has dimensions \(2.80 \mathrm{~cm}\) by \(3.20 \mathrm{~cm}\) and is in a uniform magnetic field that is directed at an angle of \(30.0^{\circ}\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of \(3.10 \times 10^{-4} \mathrm{~Wb}\) through the surface?

Short Answer

Expert verified
To get the magnitude of the magnetic field necessary to produce the given flux, perform the calculations outlined in step 5. This will get you the answer in Tesla (T), the typical unit for magnetic field strength.

Step by step solution

01

Understand Magnetic Flux

The magnetic flux through a closed surface is given by the formula: \(\Phi = B \cdot A \cdot \cos(\theta)\), where \(\Phi\) is the magnetic flux, \(B\) is the magnetic field strength, \(A\) is the area of the surface, and \(\theta\) is the angle of the magnetic field.
02

Compute the area

First, calculate the area of the surface using the given dimensions. This is done by multiplying the length and width of the surface: \(A = length \times width = 2.80 cm \times 3.20 cm = 8.96 cm^2\). Convert this to square meters: \(A = 8.96 \times 10^{-4} m^2.\)
03

Rearrange the formula for the magnetic field

To solve for the magnetic field, we need to rearrange the formula for magnetic flux: \(B = \Phi / (A \cdot \cos(\theta)).\)
04

Substitute the values

Substitute the value of the flux (\(\Phi = 3.10 \times 10^{-4} Wb\)), the area (from step 2), and the angle (\(\theta = 30.0^{\circ}\)) into the rearranged formula: \(B = (3.10 \times 10^{-4} Wb) / (8.96 \times 10^{-4} m^2 \cdot \cos(30.0^{\circ}))\).
05

Calculate the magnetic field

Now, compute the magnitude of the magnetic field by performing the above division operation, ensuring the angle is converted to radians when using the cosine function.

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Key Concepts

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

Magnetic Field Strength
Magnetic field strength, often represented by the symbol \(B\), is a measure of the intensity of a magnetic field at a particular point. It tells you how strong the magnetic influence is in a given area.
When dealing with questions about magnetic field strength, consider the context. Different environments could have varying magnetic field strengths, but the concept remains the same. It quantifies how much force would be exerted on moving charges and magnetic materials within that area. In the context of our exercise, we calculated the magnetic field strength required to produce a certain magnetic flux through the surface of a rectangle. Using the formula for magnetic flux and given values, you learn how to accurately determine the magnetic intensity needed in practical applications.
Surface Area Calculation
Surface area calculation is an essential skill in physics, particularly regarding magnetic flux problems. It's crucial because the area directly affects the magnetic flux that can pass through it.
To find the area of a rectangular surface, you multiply its length by its width. In this exercise, the surface area of the rectangle was calculated as \(2.80 \text{ cm} \times 3.20 \text{ cm} = 8.96 \text{ cm}^2\). For calculations involving magnetic flux, this area must be converted to square meters since it's a standard unit of measurement in physics. So, here the area in meters squared is \(8.96 \times 10^{-4} \text{ m}^2\). This conversion is vital for consistency in formulas used later, ensuring accurate results.
Magnetic Flux Formula
The magnetic flux formula is fundamental in determining how much of a magnetic field passes through a given area. It is expressed as:\[\Phi = B \cdot A \cdot \cos(\theta)\]\(\Phi\) represents the magnetic flux (how much magnetic field is going through the surface), \(B\) stands for magnetic field strength, \(A\) is the area through which the field lines pass, and \(\theta\) is the angle between the magnetic field and the normal (perpendicular) to the surface.
In our step-by-step solution, we arranged this formula to solve for the magnetic field \(B\). This rearrangement is helpful when you have other known quantities like flux, area, and angle. You observe how this formula integrates all these variables to ensure the magnetic field strength is tailored to produce the desired flux.
Trigonometric Functions in Physics
Trigonometric functions are invaluable tools in physics, helping to solve a range of problems. In our exercise, cosine plays a significant role in determining magnetic flux. Cosine, represented as \(\cos(\theta)\), helps account for the orientation of the magnetic field relative to a surface. It modifies the magnetic flux, depending on this angle. For instance, if the angle \(\theta\) is \(0^{\circ}\), the field is perpendicular to the surface, and \(\cos(0^{\circ}) = 1\), maximizing the flux. Conversely, if the surface is parallel (\(\theta = 90^{\circ}\)), \(\cos(90^{\circ}) = 0\), and the flux will be zero.
In this specific problem, the angle was \(30^{\circ}\), which means some, but not all, of the field lines pass through due to \(\cos(30^{\circ}) = \frac{\sqrt{3}}{2}\). This trigonometric concept shows you how orientation impacts physical phenomena like flux, illustrating the necessity of using trigonometry in physics.

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Most popular questions from this chapter

One method for determining the amount of corn in early Native American diets is the stable isotope ratio analysis (SIRA) technique. As corn photosynthesizes, it concentrates the isotope carbon-13, whereas most other plants concentrate carbon-12. Overreliance on corn consumption can then be correlated with certain diseases, because corn lacks the essential amino acid lysine. Archaeologists use a mass spectrometer to separate the \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) isotopes in samples of human remains. Suppose you use a velocity selector to obtain singly ionized (missing one electron) atoms of speed \(8.50 \mathrm{~km} / \mathrm{s}\), and you want to bend them within a uniform magnetic field in a semicircle of diameter \(25.0 \mathrm{~cm}\) for the \({ }^{12} \mathrm{C}\). The measured masses of these isotopes are \(1.99 \times 10^{-26} \mathrm{~kg}\left({ }^{12} \mathrm{C}\right)\) and \(2.16 \times 10^{-26} \mathrm{~kg}\left({ }^{13} \mathrm{C}\right) .\) (a) What strength of magnetic field is required? (b) What is the diameter of the \({ }^{13} \mathrm{C}\) semicircle? (c) What is the separation of the \({ }^{12} \mathrm{C}\) and \({ }^{13} \mathrm{C}\) ions at the detector at the end of the semicircle? Is this distance large enough to be easily observed?

You are using a type of mass spectrometer to measure charge-to-mass ratios of atomic ions. In the device, atoms are ionized with a beam of electrons to produce positive ions, which are then accelerated through a potential difference \(V\). (The final speed of the ions is great enough that you can ignore their initial speed.) The ions then enter a region in which a uniform magnetic field \(\vec{B}\) is perpendicular to the velocity of the ions and has magnitude \(B=0.250 \mathrm{~T}\). In this \(\overrightarrow{\boldsymbol{B}}\) region, the ions move in a semicircular path of radius \(R .\) You measure \(R\) as a function of the accelerating voltage \(V\) for one particular atomic ion: $$ \begin{array}{l|lllll} \boldsymbol{V}(\mathbf{k} \mathbf{V}) & 10.0 & 12.0 & 14.0 & 16.0 & 18.0 \\ \hline \boldsymbol{R}(\mathrm{cm}) & 19.9 & 21.8 & 23.6 & 25.2 & 26.8 \end{array} $$ (a) How can you plot the data points so that they will fall close to a straight line? Explain. (b) Construct the graph described in part (a). Use the slope of the best-fit straight line to calculate the charge-to-mass ratio \((q / m)\) for the ion. \((\mathrm{c})\) For \(V=20.0 \mathrm{kV},\) what is the speed of the ions as they enter the \(\vec{B}\) region? (d) If ions that have \(R=21.2 \mathrm{~cm}\) for \(V=12.0 \mathrm{kV}\) are singly ionized, what is \(R\) when \(V=12.0 \mathrm{kV}\) for ions that are doubly ionized?

A dc motor with its rotor and field coils connected in series has an internal resistance of \(3.2 \Omega .\) When the motor is running at full load on a \(120 \mathrm{~V}\) line, the emf in the rotor is \(105 \mathrm{~V}\). (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A conducting bar with mass \(m\) and length \(L\) slides over horizontal rails that are connected to a voltage source. The voltage source maintains a constant current \(I\) in the rails and bar, and a constant, uniform, vertical magnetic field \(\vec{B}\) fills the region between the rails (Fig. \(\mathbf{P 2 7 . 5 9}\) ). (a) Find the magnitude and direction of the net force on the conducting bar. Ignore friction, air resistance, and electrical resistance. (b) If the bar has mass \(m,\) find the distance \(d\) that the bar must move along the rails from rest to attain speed \(v\). (c) It has been suggested that rail guns based on this principle could accelerate payloads into earth orbit or beyond. Find the distance the bar must travel along the rails if it is to reach the escape speed for the earth \((11.2 \mathrm{~km} / \mathrm{s}) .\) Let \(B=0.80 \mathrm{~T}, I=2.0 \times 10^{3} \mathrm{~A}, m=25 \mathrm{~kg}\) and \(L=50 \mathrm{~cm} .\) For simplicity assume the net force on the object is equal to the magnetic force, as in parts (a) and (b), even though gravity plays an important role in an actual launch in space.

A particle with charge \(2.15 \mu \mathrm{C}\) and mass \(3.20 \times 10^{-11} \mathrm{~kg}\) is initially traveling in the \(+y\) -direction with a speed \(v_{0}=1.45 \times 10^{5} \mathrm{~m} / \mathrm{s} .\) It then enters a region containing a uniform magnetic field that is directed into, and perpendicular to, the page in Fig. \(\mathrm{P} 27.81 .\) The magnitude of the field is \(0.420 \mathrm{~T}\). The region extends a distance of \(25.0 \mathrm{~cm}\) along the initial direction of travel; \(75.0 \mathrm{~cm}\) from the point of entry into the magnetic field region is a wall. The length of the field-free region is thus \(50.0 \mathrm{~cm} .\) When the charged particle enters the magnetic field, it follows a curved path whose radius of curvature is \(R .\) It then leaves the magnetic field after a time \(t_{1},\) having been deflected a distance \(\Delta x_{1} .\) The particle then travels in the field-free region and strikes the wall after undergoing a total deflection \(\Delta x\). (a) Determine the radius \(R\) of the curved part of the path. (b) Determine \(t_{1}\), the time the particle spends in the magnetic field. (c) Determine \(\Delta x_{1}\), the horizontal deflection at the point of exit from the field. (d) Determine \(\Delta x\), the total horizontal deflection.
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