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Assume the average value of the vertical component of Earth's magnetic field is \(43 \mu \mathrm{T}\) (downward) for all of Arizona, which has an area of \(2.95 \times 10^{5} \mathrm{~km}^{2}\). What then are the (a) magnitude and (b) direction (inward or outward) of the net magnetic flux through the rest of Earth's surface (the entire surface excluding Arizona)?

Short Answer

Expert verified
(a) Magnitude: \(1.2685 \times 10^{7} \mathrm{~Wb}\); (b) Direction: outward.

Step by step solution

01

Understand the Given Data

We are given the average vertical component of Earth's magnetic field in Arizona as \(43 \mu \mathrm{T}\) and the area of Arizona as \(2.95 \times 10^{5} \mathrm{~km}^{2}\). The problem involves calculating the net magnetic flux through the rest of Earth's surface.
02

Calculate the Magnetic Flux through Arizona

The magnetic flux \(\Phi\) through a surface is given by the formula \(\Phi = B \cdot A\), where \(B\) is the magnetic field and \(A\) is the area of the surface. Using this formula, calculate the magnetic flux through Arizona. First, convert \(B = 43 \mu \mathrm{T}\) to Tesla: \(43 \times 10^{-6} \mathrm{~T}\). The area \(A\) in \(m^2\) is \(2.95 \times 10^{11} \mathrm{~m}^{2}\) (since \(1 \mathrm{~km}^{2} = 10^{6} \mathrm{~m}^{2}\)). Compute \(\Phi = 43 \times 10^{-6} \times 2.95 \times 10^{11} = 1.2685 \times 10^{7} \mathrm{~Wb}\) (Weber).
03

Understand the Concept of Net Flux

Under Gauss's Law for magnetism, the total magnetic flux through a closed surface is zero. This means the net magnetic flux entering through a surface must be balanced by the flux leaving it.
04

Calculate Flux through Earth's Entire Surface

Translate the law to the problem, where Earth's closed surface implies the net flux through the entire Earth's surface is zero. Thus, the flux out through the rest of the Earth's surface is equal in magnitude but opposite in direction to the flux through Arizona, since Arizona is the only area considered separately.
05

Determine Magnitude and Direction of Net Flux

Given the flux through Arizona is \(1.2685 \times 10^{7} \mathrm{~Wb}\) downward, for the rest of the Earth's surface, the same magnitude of flux must be upward. Thus, the magnitude of flux remains \(1.2685 \times 10^{7} \mathrm{~Wb}\) but is directed outward from the surface of the Earth.

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Key Concepts

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

Earth's magnetic field
Earth has a magnetic field that envelops the planet in a giant magnetic sphere. This field resembles that of a bar magnet, though its source comes from complex processes in Earth's core. The magnetic field is crucial because it protects Earth from solar and cosmic radiation.

Any point on Earth's surface can have its magnetic field described in terms of its components: horizontal, vertical, and the total field vector. The vertical component is especially important in determining how the field interacts with surfaces like Arizona's flat terrain.

This magnetic field is not uniform across the planet. Factors such as latitude and altitude can cause variations. Understanding these components helps us navigate, as they’re fundamental for compass use.
Gauss's Law for magnetism
Gauss's Law for magnetism is one of Maxwell's equations, forming the foundation of classical electromagnetism. This law states that the net magnetic flux through any closed surface is zero. In simpler terms, magnetic monopoles (isolated North or South poles) do not exist.

Because the total magnetic flux entering a closed surface must equal the flux leaving it, magnetic field lines are always continuous loops. This idea is essential when calculating flux through areas, like for Earth's regions such as Arizona and its surroundings. Gauss’s Law assures that locally measured flux can indicate what happens globally for closed surfaces like Earth.
Magnetic flux calculation
Magnetic flux quantifies the extent of a magnetic field passing through a surface. It is calculated using the equation: \[ \Phi = B \cdot A \cdot \cos(\theta) \]where \(\Phi\) is the magnetic flux, \(B\) is the magnetic field strength, \(A\) is the area through which the field lines pass, and \(\theta\) is the angle between the field lines and the normal (perpendicular) to the surface.

In problems like Arizona's flux calculation, the vertical component simplifies the equation as it implies \(\theta = 0\). Consequently, we multiply the area by the magnetic field to find the flux. This calculation reveals insights into the characteristics of the field in that specific location.
Vertical component of magnetic field
The vertical component of the magnetic field is the part that points perpendicular to Earth's surface. It significantly impacts measurements and calculations of magnetic phenomena on curved earth's surfaces.

In the case of Arizona's magnetic field, the vertical component is given as \(43 \mu \mathrm{T}\). This element specifically penetrates or exits the area directly, allowing us to compute flux without needing adjustments for angles. Understanding this component helps determine the net effect of Earth’s entire magnetic field on specific regions.

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Most popular questions from this chapter

You place a magnetic compass on a horizontal surface, allow the needle to settle, and then give the compass a gentle wiggle to cause the needle to oscillate about its equilibrium position. The oscillation frequency is \(0.312 \mathrm{~Hz}\). Earth's magnetic field at the location of the compass has a horizontal component of \(18.0 \mu \mathrm{T}\). The needle has a magnetic moment of \(0.680 \mathrm{~mJ} / \mathrm{T}\). What is the needle's rotational inertia about its (vertical) axis of rotation?

A parallel-plate capacitor with circular plates of radius \(40 \mathrm{~mm}\) is being discharged by a current of \(6.0 \mathrm{~A}\). At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to \(75 \%\) of its maximum value? (c) What is that maximum value?

A parallel-plate capacitor with circular plates of radius \(R=16 \mathrm{~mm}\) and gap width \(d=5.0 \mathrm{~mm}\) has a uniform electric field between the plates. Starting at time \(t=0,\) the potential difference between the two plates is \(V=(100 \mathrm{~V}) e^{-l / \tau},\) where the time constant \(\tau=12 \mathrm{~ms}\). At radial distance \(r=0.80 R\) from the central axis, what is the magnetic field magnitude (a) as a function of time for \(t \geq 0\) and \((\mathrm{b})\) at time \(t=3 \tau ?\)

A capacitor with parallel circular plates of radius \(R=1.20 \mathrm{~cm}\) is discharging via a current of 12.0 A. Consider a loop of radius \(R / 3\) that is centered on the central axis between the plates. (a) How much displacement current is encircled by the loop? The maximum induced magnetic field has a magnitude of \(12.0 \mathrm{mT}\). At what radius (b) inside and (c) outside the capacitor gap is the magnitude of the induced magnetic field \(3.00 \mathrm{mT} ?\)

The saturation magnetization \(M_{\max }\) of the ferromagnetic metal nickel is \(4.70 \times 10^{5} \mathrm{~A} / \mathrm{m} .\) Calculate the magnetic dipole moment of a single nickel atom. (The density of nickel is \(8.90 \mathrm{~g} / \mathrm{cm}^{3}\), and its molar mass is \(58.71 \mathrm{~g} / \mathrm{mol} .)\)

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