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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 wire \(25.0 \mathrm{~cm}\) long lies along the \(z\) -axis and carries a current of \(7.40 \mathrm{~A}\) in the \(+z\) -direction. The magnetic field is uniform and has components \(B_{x}=-0.242 \mathrm{~T}, B_{y}=-0.985 \mathrm{~T},\) and \(B_{z}=-0.336 \mathrm{~T}\) (a) Find the components of the magnetic force on the wire. (b) What is the magnitude of the net magnetic force on the wire?

A straight, \(2.5 \mathrm{~m}\) wire carries a typical household current of \(1.5 \mathrm{~A}\) (in one direction) at a location where the earth's magnetic field is 0.55 gauss from south to north. Find the magnitude and direction of the force that our planet's magnetic field exerts on this wire if it is oriented so that the current in it is running (a) from west to east, (b) vertically upward, (c) from north to south. (d) Is the magnetic force ever large enough to cause significant effects under normal household conditions?

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.

A \(150 \mathrm{~g}\) ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125 \mathrm{~m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude \(0.250 \mathrm{~T}\) and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at \(1.50 \mathrm{~km} / \mathrm{s}\) in 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 \(\begin{array}{llll}-z \text { -direction } & \text { experiences } & \text { a } & \text { force } & \text { of } & 8.50 \times 10^{-16} \mathrm{~N} & \text { in the }\end{array}\) \(+y\) -direction. (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 \(3.20 \mathrm{~km} / \mathrm{s} ?\)
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