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Chapter 32: Problem 31

In New Hampshire the average horizontal component of Earth's magnetic field in 1912 was \(16 \mu \mathrm{T},\) and the average inclination or "dip" was \(73^{\circ} .\) What was the corresponding magnitude of Earth's magnetic field?

Short Answer

Expert verified
The magnitude of Earth's magnetic field was approximately 54.63 碌T.

Step by step solution

01

Understand the Components of Earth's Magnetic Field

Earth's magnetic field can be described using its horizontal component and its dip (or inclination). The horizontal component is the magnetic field that points along the surface of the Earth. Inclination, given as an angle, is the angle made by the field with the horizontal plane. We are asked to find the magnitude of the total magnetic field using these two parts.
02

Recognize the Relationship Between Components and Magnitude

The magnitude of the Earth's magnetic field, denoted as \( B_t \,\), is related to the horizontal component \( B_h \,\) and the inclination angle \( heta \,\) by the formula \( B_t = \frac{B_h}{\cos \theta} \,\). This equation arises from the definition of the cosine of an angle in a right triangle.
03

Substitute Given Values into the Formula

We are given that the horizontal component \( B_h = 16 \, \mu T \,\) and the inclination \( \theta = 73^{\circ} \,\). Substituting these into the formula gives: \[ B_t = \frac{16 \, \mu T}{\cos 73^{\circ}}\]
04

Calculate Cosine of the Inclination Angle

Calculate \( \cos 73^{\circ} \,\) using a calculator. You find that \( \cos 73^{\circ} \approx 0.2929 \,\).
05

Compute the Magnitude of the Magnetic Field

Substitute \( \cos 73^{\circ} \approx 0.2929 \,\) into the magnitude formula to find \( B_t \,\): \[ B_t = \frac{16 \, \mu T}{0.2929} \approx 54.63 \, \mu T\]
06

Verify the Units and Result

Ensure that the result is in \( \mu T \,\), which it is. The result of \( 54.63 \, \mu T \,\) makes sense, as the total magnetic field magnitude should exceed its horizontal component due to the addition of the vertical component determined by the inclination.

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Key Concepts

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

Magnetic Field Components
Earth's magnetic field is multifaceted and can be conceptually divided into two primary components. These are the horizontal and vertical components. The horizontal component (\( B_h \)) aligns parallel to the Earth's surface, representing how the field spreads across a flat plane. Meanwhile, the vertical component is portrayed by the angle of inclination (or dip), which describes how much the field points downward into the Earth.
To visualize this, imagine standing in a landscape with invisible lines of force sweeping beneath you. The horizontal component would stretch from side to side, while the vertical component would tilt down into the ground. Together, these components represent the total behavior of Earth's magnetic field at any given location.
Inclination Angle
The inclination angle plays an important role in understanding Earth's magnetic field. It is the angle at which the magnetic field dips below the horizontal plane. Inclination is often known as the 'dip' because it describes how far the magnetic field tilts downwards.
The inclination angle ranges from \( 0^{\circ} \) at the magnetic equator, where the magnetic field is perfectly horizontal, to \( 90^{\circ} \) at the magnetic poles, where the field points directly downwards. These angles help in determining the vertical component of the magnetic field based on the known horizontal element. In simpler terms, it tells us how 'steep' or 'shallow' the field's angle is at any location on Earth, which can dramatically affect the strength and behavior of the magnetic field we observe.
Magnitude Calculation
To find the magnitude of Earth's magnetic field, we use the relationship with the horizontal component and inclination angle. The magnitude (\( B_t \)) of the total magnetic field can be calculated using the cosine function, derived from trigonometric principles in a right triangle. The formula is: \[ B_t = \frac{B_h}{\cos \theta} \], where \( \theta \) is the inclination angle.
This equation helps us translate the horizontal component into the total field's magnitude by considering how 'tilted' the field lines are. Calculating \( \cos 73^{\circ} \) gives us an understanding of how much the horizontal and vertical components contribute to the overall magnitude. Once we substitute these values, we arrive at the result \( 54.63 \, \mu T \). This makes sense because the total field is naturally greater than just the horizontal part due to the inclination's vertical component.

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Most popular questions from this chapter

A parallel-plate capacitor with circular plates is being charged. Consider a circular loop centered on the central axis and located between the plates. If the loop radius of \(3.00 \mathrm{~cm}\) is greater than the plate radius, what is the displacement current between the plates when the magnetic field along the loop has magnitude \(2.00 \mu \mathrm{T} ?\)

A parallel-plate capacitor with circular plates of radius \(0.10 \mathrm{~m}\) is being discharged. A circular loop of radius \(0.20 \mathrm{~m}\) is concentric with the capacitor and halfway between the plates. The displacement current through the loop is 2.0 A. At what rate is the electric field between the plates changing?

The magnitude of the magnetic dipole moment of Earth is \(8.0 \times 10^{22} \mathrm{~J} / \mathrm{T}\). (a) If the origin of this magnetism were a magnetized iron sphere at the center of Earth, what would be its radius? (b) What fraction of the volume of Earth would such a sphere occupy? Assume complete alignment of the dipoles. The density of Earth's inner core is \(14 \mathrm{~g} / \mathrm{cm}^{3} .\) The magnetic dipole moment of an iron atom is \(2.1 \times 10^{-23} \mathrm{~J} / \mathrm{T}\). (Note: Earth's inner core is in fact thought to be in both liquid and solid forms and partly iron, but a permanent magnet as the source of Earth's magnetism has been ruled out by several considerations. For one, the temperature is certainly above the Curie point.)

An electron is placed in a magnetic field \(\vec{B}\) that is directed along a \(z\) axis. The energy difference between parallel and antiparallel alignments of the \(z\) component of the electron's spin magnetic moment with \(\vec{B}\) is \(6.00 \times 10^{-25} \mathrm{~J}\). What is the magnitude of \(\vec{B} ?\)

Suppose that a parallel-plate capacitor has circular plates with radius \(R=30 \mathrm{~mm}\) and a plate separation of \(5.0 \mathrm{~mm}\). Suppose also that a sinusoidal potential difference with a maximum value of \(150 \mathrm{~V}\) and a frequency of \(60 \mathrm{~Hz}\) is applied across the plates; that is, $$V=(150 \mathrm{~V}) \sin [2 \pi(60 \mathrm{~Hz}) t]$$ (a) Find \(B_{\max }(R),\) the maximum value of the induced magnetic field that occurs at \(r=R .\) (b) Plot \(B_{\max }(r)\) for \(0
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