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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) What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56 \(\times\) 10\(^4\) V/m and a magnetic field of 4.62 \(\times\) 10\(^{-3}\) T, with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\vec{v}\), \(\overrightarrow{E}\), and \(\overrightarrow{B}\). (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

A proton (\(q\) = 1.60 \(\times\) 10\(^{-19}\) C, \(m =\) 1.67 \(\times\) 10\(^{-27}\) kg) moves in a uniform magnetic field \(\overrightarrow{B} =\) (0.500 T)\(\hat{\imath}\). At \(t =\) 0 the proton has velocity components \(\upsilon_x =\) 1.50 \(\times\) 10\(^5\) m/s, \(\upsilon_y =\) 0, and \(\upsilon_z =\) 2.00 \(\times\) 10\(^5\) m/s (see Example 27.4). (a) What are the magnitude and direction of the magnetic force acting on the proton? In addition to the magnetic field there is a uniform electric field in the +\(x\)-direction, \(\overrightarrow{E} =\) (+2.00 \(\times\) 10\(^4\) V/m)\(\hat{\imath}\). (b) Will the proton have a component of acceleration in the direction of the electric field? (c) Describe the path of the proton. Does the electric field affect the radius of the helix? Explain. (d) At \(t =\) \(T\)/2, where T is the period of the circular motion of the proton, what is the \(x\)-component of the displacement of the proton from its position at \(t =\) 0?

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of 2.00 kV. Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 m. What is the magnitude of the field?

The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes \(^{15}\)N to \(^{14}\)N in bone from human remains. Carnivores concentrate \(^{15}\)N, so this ratio tells archaeologists how much meat was consumed. For a mass spectrometer that has a path radius of 12.5 cm for \(^{12}\)C ions (mass 1.99 \(\times\) 10\(^{-26}\) kg), find the separation of the \(^{14}\)N 1mass 2.32 \(\times\) 10\(^{-26}\) kg2 and 15N (mass 2.49 \(\times\) 10\(^{-26}\) kg) isotopes at the detector.

If two deuterium nuclei (charge \(+e\), mass 3.34 \(\times\) 10\(^{-27}\) kg) get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 10\(^{-15}\) m. This is the principle behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained by physical walls, so they are confined magnetically. (a) How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Assume their speeds are equal. Treat the nuclei as point charges, and assume that a separation of 1.0 \(\times\) 10\(^{-15}\) is required for fusion.) (b) What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.50 m?

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