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Chapter 27: Problem 10

A flat, square surface with side length \(3.40 \mathrm{~cm}\) is in the \(x y\) -plane at \(z=0 .\) Calculate the magnitude of the flux through this surface produced by a magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.200 \mathrm{~T}) \hat{\imath}+(0.300 \mathrm{~T}) \hat{\jmath}-(0.500 \mathrm{~T}) \hat{\boldsymbol{k}}\).

Short Answer

Expert verified
The magnetic flux through the surface is -0.000578 Weber.

Step by step solution

01

Identify the relevant variables

The magnetic field is \( \overrightarrow{B}=(0.200 \, \mathrm{T}) \, \hat{i} +(0.300 \, \mathrm{T}) \, \hat{j} -(0.500 \, \mathrm{T}) \, \hat{k} \), and the side length of the square surface is \( 3.40 \, \mathrm{cm} = 0.034 \, \mathrm{m} \). We'll use these values in the formula for magnetic flux.
02

Calculate the area of the square

The area of the surface (\( A \)) is the square of the side length: \( A = (0.034 \, \mathrm{m})^2 = 0.001156 \, \mathrm{m}^2 \).
03

Compute the magnetic flux

Substitute the calculated area and the k-component of the B-field into the formula for magnetic flux: \( \Phi = \overrightarrow{B} \cdot \overrightarrow{A} = B_k \times A = -0.500 \, \mathrm{T} \times 0.001156 \, \mathrm{m}^2 = -0.000578 \, \mathrm{T} \cdot \mathrm{m}^2 = -0.000578 \, \mathrm{Wb} \).
04

Interpret the result

The negative sign indicates that the direction of the flux is downward, or in the negative z-direction, which is expected since we chose the upwards direction (along the positive z-axis) as positive by convention.

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Key Concepts

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

Magnetic Field Vector
The magnetic field vector, often represented by \( \overrightarrow{B} \) or simply \( B \) with an arrow to denote its vector nature, is an essential concept in electromagnetic theory. This vector describes the direction and magnitude of the magnetic field at a particular point in space. For example, in the given exercise, the field is defined as \( \overrightarrow{B} = (0.200 \, \mathrm{T}) \, \hat{i} +(0.300 \, \mathrm{T}) \, \hat{j} -(0.500 \, \mathrm{T}) \, \hat{k} \).

This notation indicates that the magnetic field has components in the x, y, and z directions of a Cartesian coordinate system. The unit of measurement for the magnetic field is the Tesla \( (T) \), representing a significant level of magnetic field strength. Understanding each component's contribution, as we can see here, is critical when calculating related properties like magnetic flux.
Flux Calculation
The calculation of magnetic flux \( (\Phi) \) entails evaluating how much magnetic field passes through a given area. To calculate the flux, we use the formula \( \Phi = \overrightarrow{B} \cdot \overrightarrow{A} \), where \( \overrightarrow{A} \) is the area vector perpendicular to the surface, and the dot product represents the multiplication of the relevant magnetic field component and the area. When dealing with a flat surface aligned along the plane, such as in our exercise, only the magnetic field's component perpendicular to the surface contributes to the flux.

Thus, for our exercise, we isolated the z-component (negative in this case) of the magnetic field vector and computed the flux with the given area of the square surface. This yielded \( \Phi = -0.000578 \, \mathrm{Wb} \), where \( \mathrm{Wb} \) (Weber) is the unit of magnetic flux. The negative result implies opposition to the arbitrarily chosen positive direction, aligning with conventions in vector analysis.
Area Vector
The area vector \( \overrightarrow{A} \) represents not just the size, but also the orientation of a given surface. It is always perpendicular to the flat surface and has a magnitude equal to the area of that surface. To establish this vector, we need both the magnitude, calculated as the area \( (A) \) of the surface, and the direction, generally taken to be normal (perpendicular) to the surface.

In our problem, the square surface is in the xy-plane at \( z=0 \) which means that the area vector points in the \( +z \) or \( -z \) direction, depending on the chosen convention. For our square surface with side length \( 3.40 \, \mathrm{cm} \) or \( 0.034 \, \mathrm{m} \), the area was calculated and found to be \( 0.001156 \, \mathrm{m}^2 \). The orientation of \( \overrightarrow{A} \) along with the z-axis component of \( \overrightarrow{B} \) is then used to find the magnetic flux through the surface.

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Most popular questions from this chapter

A small particle with positive charge \(q=+3.75 \times 10^{-4} \mathrm{C}\) and mass \(m=5.00 \times 10^{-5} \mathrm{~kg}\) is moving in a region of uniform electric and magnetic fields. The magnetic field is \(B=4.00 \mathrm{~T}\) in the \(+z\) -direction. The electric field is also in the \(+z\) -direction and has magnitude \(E=60.0 \mathrm{~N} / \mathrm{C}\). At time \(t=0\) the particle is on the \(y\) -axis at \(y=+1.00 \mathrm{~m}\) and has velocity \(v=30.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction. Neglect gravity. (a) What are the \(x\) -, \(y\) and \(z\) -coordinates of the particle at \(t=0.0200 \mathrm{~s} ?\) (b) What is the speed of the particle at \(t=0.0200 \mathrm{~s} ?\)

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of \(6.64 \times 10^{-27} \mathrm{~kg}\) ) traveling horizontally at \(35.6 \mathrm{~km} / \mathrm{s}\) enters a uniform, vertical, \(1.80 \mathrm{~T}\) magnetic field. (a) What is the diameter of the path followed by this alpha particle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.

A particle with charge \(q\) and mass \(m\) is dropped at time \(t=0\) from rest at its origin in a region of constant magnetic field \(\overrightarrow{\boldsymbol{B}}\) that points horizontally. What happens? To answer, construct a Cartesian coordinate system with the \(y\) -axis pointing downward and the \(z\) -axis pointing in the direction of the magnetic field. At time \(t \geq 0\) the particle has velocity \(\overrightarrow{\boldsymbol{v}}=v_{x} \hat{\imath}+v_{y} \hat{\jmath} .\) The net force \(\overrightarrow{\boldsymbol{F}}=F_{x} \hat{\imath}+F_{y} \hat{\jmath}\) on the particle is the vector sum of its weight and the magnetic force. (a) Using Newton's second law, write equations for \(a_{x}\) and \(a_{y},\) where \(\vec{a}=a_{x} \hat{\imath}+a_{y} \hat{\jmath}\) is the acceleration of the particle. (b) Differentiate the second of these equations with respect to time. Then substitute your expression for \(a_{x}=d v_{x} / d t\) to determine an equation for \(d v_{y}^{2} / d t^{2}\) in terms of \(v_{y}\) (c) This result shows that \(v_{y}\) is a simple harmonic oscillator. Use the initial conditions to determine \(v_{y}(t) .\) Write your answer in terms of the angular frequency \(\omega=q B / m\). Note that \(\left(d v_{y} / d t\right)_{0}=g .\) (d) Substitute your result for \(v_{y}(t)\) into your equation for \(d v_{x} / d t .\) Integrate using the initial conditions to determine \(v_{x}(t) .\) (e) Integrate your expressions for \(v_{x}(t)\) and \(v_{y}(t)\) to determine \(x(t)\) and \(y(t) .\) (f) If \(m=1.00 \mathrm{mg}, q=19.6 \mu \mathrm{C}\) and \(B=10.0 \mathrm{~T}\), what maximum vertical distance does the particle drop before returning upward?

A particle with charge \(-5.60 \mathrm{nC}\) is moving in a uniform magnetic field \(\vec{B}=-(1.25 \mathrm{~T}) \hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{\boldsymbol{F}}=-\left(3.40 \times 10^{-7} \mathrm{~N}\right) \hat{\imath}+\left(7.40 \times 10^{-7} \mathrm{~N}\right) \hat{\jmath}\) (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 \(\vec{v} \cdot \vec{F}\). What is the angle between \(\vec{v}\) and \(\vec{F} ?\)

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?
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