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Assume that the Earth's magnetic field is a dipole. At what distance above the Earth's surface is the magnitude of the field one-half of its value at the surface?

Short Answer

Expert verified
The distance is approximately 1,003 km above Earth's surface.

Step by step solution

01

Understanding the Earth's Magnetic Field

The Earth's magnetic field can be approximated as a dipole field. The magnetic field strength, at a distance \( r \) from the center of the Earth, is proportional to \( rac{1}{r^3} \). We want to find the distance \( h \) above Earth's surface where the magnetic field strength is half of its value at the surface.
02

Defining the Relationship at Earth's Surface

Let \( B_s \) be the magnetic field strength at Earth's surface. At the surface, the distance from the Earth's center is \( R \), where \( R \) is the radius of the Earth. Therefore, \( B_s \propto \frac{1}{R^3} \).
03

Expressing the Desired Magnetic Field

Let \( B_h \) be the magnetic field strength at height \( h \) above the surface, where \( B_h = \frac{1}{2}B_s \). At this height, the distance from Earth's center is \( R + h \), so \( B_h \propto \frac{1}{(R+h)^3} \).
04

Solving for the Desired Height

Setting the equation for magnetic field strength, we have:\[ \frac{1}{2}B_s = rac{B_s}{(1+h/R)^3} \]This simplifies to:\[ rac{1}{2} = rac{1}{(1+h/R)^3} \]Rearrange to solve for \( h \):\[ (1+h/R)^3 = 2 \]
05

Calculating Height Above the Surface

Take the cube root of both sides:\[ 1+h/R = 2^{1/3} \]Solving for \( h \), we have:\[ h = R(2^{1/3} - 1) \]Substitute Earth's radius, \( R \approx 6,371 \, \text{km} \), into the equation:\[ h = 6,371 \, \text{km} \times (2^{1/3} - 1) \approx 1,003 \, \text{km} \]

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Key Concepts

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

Magnetic Dipole
The concept of a magnetic dipole is fundamental when describing the Earth's magnetic field. A magnetic dipole consists of two opposite magnetic charges or poles separated by a distance. Imagine a simple bar magnet with a north and south pole; this is a classic example of a dipole. The Earth itself behaves like a giant magnetic dipole due to its inner core and the motion of molten iron within it.

When we refer to the Earth as a dipole, we simplify the complex magnetic field into a more manageable two-pole system. This simplification proves crucial for calculations involving the field at various distances from the Earth's surface. It allows us to predict how the field strength changes as we move away from the Earth, following specific mathematical relationships.
Field Strength
Field strength in the context of Earth's magnetic field refers to the intensity of the magnetic effect at a particular point in space. It indicates how strong the magnetic influence is and can be measured in units called teslas.

For a magnetic dipole like the Earth, the strength of the magnetic field decreases with distance. Mathematically, this relationship is expressed with the rule that field strength is inversely proportional to the cube of the distance from the magnetic source. In equations, this is shown as \( B \propto \frac{1}{r^3} \), where \( B \) represents the field strength and \( r \) is the distance from the center of the Earth. This provides a clear understanding that as you move further from the source, in this case, the Earth, the magnetic influence weakens considerably.

Knowing the expression of field strength relative to distance allows us to solve problems, like determining where the field strength becomes a fraction of its original value at the surface.
Distance Calculation
Determining the specific distance where the magnetic field strength becomes a fraction of its initial value at the surface requires understanding this distance calculation. It is particularly useful in applications like satellite technology, where field variations are quite significant.

To solve such problems, we set up equations based on the knowledge that field strength diminishes with the cube of the increase in distance. For instance, when seeking the height above Earth's surface where the field is half its surface strength, we use the equation: \( \frac{1}{2}B_s = \frac{B_s}{(1+h/R)^3} \). Solving this, we find \( h \), the height above the surface.

The solution involves algebraic manipulation, taking the cube root of the ratio expression, ultimately resulting in a calculation where \( h = R(2^{1/3} - 1) \). This formula beautifully demonstrates the power of mathematical relationships in physics.
Radius of the Earth
The radius of the Earth is a crucial value in computations involving the Earth's magnetic field. The average radius, which is approximately 6,371 kilometers, serves as a baseline measurement for defining distances from the center of the Earth.

This measurement becomes particularly relevant in scenarios requiring distance-based computations, like when calculating the height above the Earth's surface where a specific magnetic field strength is observed. In such cases, the radius is essential to establish initial conditions and relationships for the calculations.

By knowing \( R \), not only can we determine magnetic field variations at specific heights, but it also allows us to appreciate the scale and scope of the Earth's magnetic dominance over its surrounding space.

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Most popular questions from this chapter

The measured declination and inclination of the paleomagnetic field in Upper Triassic rocks at \(41.5^{\circ} \mathrm{N}\) and \(72.7^{\circ} \mathrm{W}\) are \(D=18^{\circ}\) and \(I=12^{\circ} .\) Deter- mine the paleomagnetic pole position. A MATLAB code for solving this problem is given in Appendix \(D\).

The measured declination and inclination of the paleomagnetic field in Oligocene rocks at \(51^{\circ} \mathrm{N}\) and \(14.7^{\circ} \mathrm{E}\) are \(D=200^{\circ}\) and \(I=-63^{\circ} .\) Determine the paleomagnetic pole position. A MATLAB code for solving this problem is given in Appendix \(D\).

At what depth will ascending mantle rock with a temperature of \(1600 \mathrm{~K}\) melt if the equation for the solidus temperature \(T\) is $$ T(K)=1500+0.12 p(\mathrm{MPa}) $$ Assume \(\rho=3300 \mathrm{~kg} \mathrm{~m}^{-3}, g=10 \mathrm{~m} \mathrm{~s}^{-2},\) and the mantle rock ascends at constant temperature.

What is the relative plate velocity between the Nazca and South American plates at Lima, Peru \(\left(12^{\circ} \mathrm{S},\right.\) \(\left.77^{\circ} \mathrm{W}\right) ?\) A MATLAB code for solving this problem is given in Appendix \(D\).

If we assume that the current rate of subduction, \(0.09 \mathrm{~m}^{2} \mathrm{~s}^{-1}\), has been applicable in the past, what thickness of sediments would have to have been subducted in the last 3 Gyr if the mass of subducted sediments is equal to one-half the present mass of the continents? Assume the density of the continents \(\rho_{c}\) is \(2700 \mathrm{~kg} \mathrm{~m}^{-3},\) the density of the sediments \(\rho_{s}\) is \(2400 \mathrm{~kg} \mathrm{~m}^{-3},\) the continental area \(A_{c}\) is \(1.9 \times 10^{8} \mathrm{~km}^{2},\) and the mean continental thickness \(h_{c}\) is \(35 \mathrm{~km}\). A MATLAB code for the solution of this problem is provided in Appendix \(D\). An introduction to the use of MATLAB is given in Section 11.2 .

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