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

Short Answer

Expert verified
The magnitude of the magnetic flux is \(0.578 \times 10^{-3}\) Weber.

Step by step solution

01

Understand the Problem

We need to calculate the magnetic flux through a square surface with side 3.40 cm. The magnetic field vector is given as \(\overrightarrow{B} = (0.200) \hat{\imath} + (0.300) \hat{\jmath} - (0.500) \hat{k}\), and the square lies in the xy-plane, making its normal vector \(\hat{k}\).
02

Calculate the Area of the Square

The side length of the square is 3.40 cm. Converting it to meters gives us \(3.40 \times 10^{-2}\) m. The area \(A\) of the square is given by:\[A = (3.40 \times 10^{-2} \text{ m})^2 = 1.156 \times 10^{-3} \text{ m}^2\]
03

Determine the Relevant Component of Magnetic Field

Since the surface is on the xy-plane, the normal vector is \(\hat{k}\). Only the component of \(\overrightarrow{B}\) along \(\hat{k}\) contributes to the flux. Thus, the relevant component is: \[-0.500 \text{ T} \hat{k}\]
04

Calculate the Magnetic Flux

Magnetic flux \(\Phi_B\) through the surface is given by the dot product of the magnetic field and the area vector \(\overrightarrow{A}\), which points in the \(\hat{k}\) direction: \[\Phi_B = \overrightarrow{B} \cdot \overrightarrow{A} = (0.200 \hat{\imath} + 0.300 \hat{\jmath} - 0.500 \hat{k}) \cdot (1.156 \times 10^{-3} \hat{k}) = -0.500 \times 1.156 \times 10^{-3} \]\[\Phi_B = -0.578 \times 10^{-3} \text{ Wb}\]
05

Magnitude of the Flux

The magnitude of the flux is the absolute value of the calculated flux, which is:\[|\Phi_B| = 0.578 \times 10^{-3} \text{ Wb}\]

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Key Concepts

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

Understanding Magnetic Fields
Magnetic fields are invisible forces that exert influence on moving electric charges, magnetic material, and currents. These fields are represented by vector quantities, meaning they have both magnitude and direction.
In the exercise, the magnetic field \(\overrightarrow{B}\) is given in terms of its components along the x, y, and z axes: \(\overrightarrow{B} = (0.200) \hat{\imath} + (0.300) \hat{\jmath} - (0.500) \hat{k}\). This means:
  • 0.200 Tesla in the x-direction (\(\hat{\imath}\)),
  • 0.300 Tesla in the y-direction (\(\hat{\jmath}\)), and
  • -0.500 Tesla in the z-direction (\(\hat{k}\)).
The negative sign indicates the direction of the field component is opposite to the positive axis.
Understanding the vector form of a magnetic field is essential for solving problems that involve magnetic flux, as it determines which component influences the flux through a given surface.
Flux Calculation Basics
Magnetic flux indicates how much of the magnetic field passes through a given surface. The key to calculating magnetic flux is using the dot product of the magnetic field vector \(\overrightarrow{B}\) and the area vector \(\overrightarrow{A}\).
The equation for magnetic flux \(\Phi_B\) is: \[\Phi_B = \overrightarrow{B} \cdot \overrightarrow{A}\] Where:
  • \(\overrightarrow{B}\) is the magnetic field vector,
  • \(\overrightarrow{A}\) is the area vector which is perpendicular to the surface,
  • The \(\cdot\) represents the dot product, which multiples magnitudes of the vectors and the cosine of the angle between them.
For a surface oriented in the xy-plane, only the z-component of \(\overrightarrow{B}\) (i.e., along \(\hat{k}\)) would contribute to the flux since the area vector points in that direction.
By aligning the surface area vector correctly with the magnetic field vector, the flux is simplified to: \[\Phi_B = B_z A\] where \(B_z\) is the z-component of the magnetic field.
Role of Surface Area in Flux
The surface area plays a crucial role in determining how much of the magnetic field penetrates it. For a flat surface, such as a square, the area vector points perpendicular to the plane of the square.
In the given exercise, the square lies in the xy-plane, making the normal vector point in the z-direction. This means it only intercepts the magnetic field component aligned with this normal vector.
The area \(A\) of the square is calculated using the formula: \[A = \text{side length}^2\] Since the side of the square is given in centimeters, converting it to meters is necessary to keep it consistent with other units.
  • First convert side length: \(3.40 \times 10^{-2} \text{ m}\)
  • Calculate area: \[A = (3.40 \times 10^{-2} \text{ m})^2 = 1.156 \times 10^{-3} \text{ m}^2\]
This calculated area is then used to find the magnetic flux through the surface. The correct measurement of surface area is essential for obtaining accurate flux values.
Applying Vector Analysis in Flux Problems
Vector analysis simplifies understanding physical phenomena involving directions and magnitudes, like magnetic fields and flux. A fundamental aspect of vector analysis in flux calculations is understanding dot products.
The dot product \(\overrightarrow{B} \cdot \overrightarrow{A}\) simplifies to: \[B_x A_x + B_y A_y + B_z A_z\] Each vector component is multiplied and summed to give a scalar quantity representing the flux through the surface.
  • Since \(\overrightarrow{A}\) is perpendicular to the plane, its only non-zero component is along \(\hat{k}\),
  • Thus, the flux is influenced only by the \(\hat{k}\) component of \(\overrightarrow{B}\), reducing the equation to: \[\Phi_B = B_z A_z\]
Understanding the role of each vector component and how the dot product helps align the vectors correctly is crucial. It enables us to focus on the relevant components affecting the magnetic flux through a surface.
Efficiently using vector analysis can make solving complex physics problems straightforward and logical.

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Most popular questions from this chapter

A mass spectrograph is used to measure the masses of ions, or to separate ions of different masses (see Section 27.5). In one design for such an instrument, ions with mass \(m\) and charge \(q\) are accelerated through a potential difference \(V\). They then enter a uniform magnetic field that is perpendicular to their velocity, and they are deflected in a semicircular path of radius \(R\). A detector measures where the ions complete the semicircle and from this it is easy to calculate \(R\). (a) Derive the equation for calculating the mass of the ion from measurements of \(B\), \(V\), \(R\), and \(q\). (b) What potential difference \(V\) is needed so that singly ionized \(^{12}\)C atoms will have \(R =\) 50.0 cm in a 0.150-T magnetic field? (c) Suppose the beam consists of a mixture of \(^{12}\)C and \(^{14}\)C ions. If \(v\) and \(B\) have the same values as in part (b), calculate the separation of these two isotopes at the detector. Do you think that this beam separation is sufficient for the two ions to be distinguished? (Make the assumption described in Problem 27.59 for the masses of the ions.)

A circular loop of wire with area \(A\) lies in the \(xy\)-plane. As viewed along the \(z\)-axis looking in the -\(z\)-direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{B}\) is given by \(\vec{\tau}\) = D(4\(\hat{\imath}\) - 3\(\hat{\jmath}\)), where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = -\vec{\mu}\) \(\cdot\) \(\overrightarrow{B}\) is negative. The magnitude of the magnetic field is \(B_0 = 13D/IA\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_x\), \(B_y\), and \(B_z\) of \(\overrightarrow{B}\).

A horizontal rectangular surface has dimensions 2.80 cm by 3.20 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\(^{-4}\) Wb through the surface?

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 \(\overrightarrow{B}\) fills the region between the rails (\(\textbf{Fig. P27.65}\)). (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 km/s). Let \(B =\) 0.80 T, \(I =\) 2.0 \(\times\) 10\(^3\) A, \(m =\) 25 kg, and \(L =\) 50 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.

An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 \(\times\) 10\(^{-27}\) kg) traveling horizontally at 35.6 km>s enters a uniform, vertical, 1.80-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.

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