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Chapter 27: Problem 10

The magnetic flux through one face of a cube is \(+0.120 \mathrm{Wb} .\) (a) What must the total magnetic flux through the other five faces of the cube be? (b) Why didn't you need to know the dimensions of the cube in order to answer part (a) 2 (c) Suppose the magnetic flux is due to a permanent magnet like that shown in Fig. \(27.11 .\) In a sketch, show where the cube in part (a) might be located relative to the magnet.

Short Answer

Expert verified
(a) -0.120 Wb; (b) Gauss's Law; (c) Near the north pole of the magnet.

Step by step solution

01

Understanding Magnetic Flux and Gauss's Law

In a closed surface, the net magnetic flux is always zero. This is Gauss's Law for magnetism, which states that the total magnetic flux through any closed surface is zero. This implies that the magnetic field lines entering a closed surface must equal the lines exiting it.
02

Applying Gauss's Law to the Cube

For the cube, the net magnetic flux through all six faces must add up to zero. Given one face has a flux of \(+0.120\, \mathrm{Wb}\), the sum of the flux through the other five faces must be \(-0.120\, \mathrm{Wb}\) to ensure the total flux is zero.
03

Non-reliance on Cube Dimensions

Gauss's Law for magnetism doesn't depend on the size or shape of the object. It applies equally to any closed surface regardless of its dimensions. Thus, the dimensions of the cube are irrelevant for calculating total magnetic flux.
04

Placing the Cube Relative to the Magnet

To sketch the cube's placement, consider that magnetic field lines exit from the north pole and enter the south pole of a magnet. Position the cube such that its face with positive flux aligns with exiting field lines. The sketch would show the cube near the north pole of the magnet, aligning one face with the exiting lines.

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Key Concepts

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

Magnetic Flux
Magnetic flux is a measure of the quantity of magnetism, taking into account the strength and extent of a magnetic field. It is denoted in Weber (Wb).
Magnetic flux through a surface is calculated by multiplying the magnetic field strength (\( B \) ) by the area (\( A \) ) the field penetrates, and the angle (\( \theta \) ) between the field and the normal (perpendicular) to the surface: \(\Phi = B \times A \times \cos(\theta)\).
This concept is important because a change in magnetic flux can induce an electromotive force (emf) in nearby conductors, which is the principle behind many electrical generators and transformers. In the exercise, the cube has six faces, and the magnetic flux through one of these faces is given as \( +0.120 \text{ Wb} \). The task is to calculate the total magnetic flux through the other five faces.
Closed Surface
A closed surface is a surface that completely encloses a space, much like a sphere or a cube. In the context of Gauss's Law for magnetism, it is crucial because it clarifies how magnetic fields behave.
According to Gauss's Law for Magnetism, the net magnetic flux through a closed surface is always zero. This means that the total magnetic flux entering the surface equals the flux exiting the surface.
This concept is helpful to answer question (a) in the exercise, where, given a flux through one face of the cube, the flux through the remaining five faces must cumulatively negate it to satisfy Gauss's Law, resulting in a net flux of zero through the cube.
Cube Placement
The placement of the cube in magnetic fields is influenced by how magnetic flux enters and exits the magnet. When dealing with a permanent magnet, field lines emerge from the north pole and loop around to enter the south pole. The exercise hints at placing the cube such that one face receiving \( +0.120 ext{ Wb} \) from the magnetic field might be aligned with these field lines exiting from the north pole.
By understanding the distribution of magnetic field lines around a magnet, we can position the cube to appropriately line up one face for optimal observation of magnetic flux effects. This mental model assists in visualizing the roles of different cube faces and how magnetic fields interact with them.

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Most popular questions from this chapter

A proton \(\left(q=1.60 \times 10^{-19} \mathrm{C}, m=1.67 \times 10^{-27} \mathrm{kg}\right)\) moves in a uniform magnetic field \(\overrightarrow{\boldsymbol{B}}=(0.500 \mathrm{T}) \hat{\boldsymbol{i}} .\) At \(t=0\) the proton has velocity components \(v_{x}=1.50 \times 10^{5} \mathrm{m} / \mathrm{s}, v_{y}=0,\) and \(v_{z}=2.00 \times 10^{5} \mathrm{m} / \mathrm{s}(\text { see Example } 27.4) .\) (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, \(\vec{E}=\left(+2.00 \times 10^{4} \mathrm{V} / \mathrm{m}\right) \hat{\imath}\) (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 \(t=0 ?\)

A particle with charge \(6.40 \times 10^{-19} \mathrm{C}\) travels in a circular orbit with radius 4.68 \(\mathrm{mm}\) due to the force exerted on it by a magnetic field with magnitude 1.65 \(\mathrm{T}\) and perpendicular to the orbit. (a) What is the magnitude of the linear momentum \(\vec{p}\) of the partcle? (b) What is the magnitude of the angular momentum \(\overrightarrow{\boldsymbol{L}}\) of the particle?

A particle of mass 0.195 g carries a charge of \(-2.50 \times\) \(10^{-8} \mathrm{C}\) . The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

A wire 25.0 \(\mathrm{cm}\) long lies along the \(z\) -axis and carries a current of 9.00 \(\mathrm{A}\) 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?

An insulated wire with mass \(m=5.40 \times 10^{-5} \mathrm{kg}\) is bent into the shape of an inverted U such that the horizontal part has a length \(l=15.0 \mathrm{cm} .\) The bent ends of the wire are partially immersed in two pools of mercury, with 2.5 \(\mathrm{cm}\) of each end below the mercury's surface. The entire structure is in a region containing a uniform \(0.00650-\mathrm{T}\) magnetic field directed into the page (Fig. 27.71\()\) . An electrical connection from the mercury pools is made through the ends of the wires. The mercury pools are connected to a \(1.50-\mathrm{V}\) battery and a switch \(\mathrm{S}\) . When switch \(\mathrm{S}\) is closed, the wire jumps 35.0 \(\mathrm{cm}\) into the air, measured from its initial position. (a) Determine the speed \(v\) of the wire as it leaves the mercury. (b) Assuming that the current \(I\) through the wire was constant from the time the switch was closed until the wire left the mercury, determine \(I\) (c) Ignoring the resistance of the mercury and the circuit wires, determine the resistance of the moving wire.
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