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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 54.79 \(\mu\text{T}\).

Step by step solution

01

Understand the Components of Earth's Magnetic Field

Earth's magnetic field can be broken down into two main components: the horizontal component and the vertical component (which depends on the dip angle). Given that the horizontal component is 16 \( \mu \text{T} \) and the dip angle is \( 73^{\circ} \), we can use these to find the magnitude of the total magnetic field.
02

Identify the Relationship

The total magnitude of Earth's magnetic field \( B \) can be found using the horizontal component \( B_h \) and the angle of inclination \( \theta \) through the relation: \( B_h = B \cos(\theta) \). Rearranging for \( B \), we have \( B = \frac{B_h}{\cos(\theta)} \).
03

Calculate the Magnitude of the Magnetic Field

Substitute the given values into the formula. Use \( B_h = 16 \mu \text{T} \) and \( \theta = 73^{\circ} \): \[ B = \frac{16 \mu \text{T}}{\cos(73^{\circ})} \]Next, calculate \( \cos(73^{\circ}) \), which is approximately 0.292, and substitute it into the equation:\[ B = \frac{16 \mu \text{T}}{0.292} \approx 54.79 \mu \text{T} \].

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Key Concepts

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

Horizontal Component
The horizontal component of Earth's magnetic field is a crucial measurement. It represents the part of the magnetic field that lies parallel to the Earth's surface. It affects navigation systems and compass readings because this component aligns with the magnetic north. For example, if the horizontal component is known, it provides direct information about how a compass needle will behave.
  • This component is typically measured in microteslas (\( \mu \text{T} \)).
  • It remains significant in geophysical surveys and studies.
Understanding the horizontal component helps in calculating the total intensity of the Earth's magnetic field when combined with other factors, such as the inclination angle.
Inclination Angle
The inclination angle, also known as the magnetic dip, refers to the angle made by the Earth's magnetic field with the horizontal surface. This tilt of the magnetic field line varies depending on the location.
  • In New Hampshire, as mentioned in our example, the inclination angle is \( 73^{\circ} \).
  • A steep inclination indicates that the magnetic field lines are dipping steeply towards the Earth鈥檚 core.
The inclination angle is key to understanding how deep the magnetic field lines dive into the Earth, which influences their strength and orientation. Understanding this angle's value is necessary for determining other properties of the Earth's magnetic field.
Magnetic Dip
Magnetic dip is closely related to the inclination angle, defined as the angle that the magnetic field makes with the line through the horizontal plane. It provides insights into how magnetic instruments need to be adjusted to account for these variations.
  • Magnetic dip varies from 0掳 at the magnetic equator to 90掳 at the magnetic poles.
  • It affects the angle at which magnetic forces act on objects on Earth's surface.
Engineers and scientists often consider the magnetic dip when planning and executing activities that rely on magnetic fields, such as in the calibration of navigational equipment.
Magnetic Field Magnitude
The magnetic field magnitude refers to the overall strength of the Earth's magnetic field. It is calculated by considering both horizontal and vertical components, suitably combined using mathematical equations. The total magnetic field magnitude is pivotal for various scientific applications.To find it, use the formula:\[ B = \frac{B_h}{\cos(\theta)} \], where:
  • \( B \) is the total magnetic field magnitude.
  • \( B_h \) is the horizontal component.
  • \( \theta \) is the inclination angle.
Using our example, with \( B_h = 16 \mu \text{T} \) and \( \theta = 73^{\circ} \), we calculate \( B \approx 54.79 \mu \text{T} \). This calculation shows the power and reach of Earth's magnetic influence at that location.

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Most popular questions from this chapter

Prove that the displacement current in a parallel-plate capacitor of capacitance \(C\) can be written as \(i_{d}=C(d V / d t)\), where \(V\) is the potential difference between the plates.

The exchange coupling mentioned in Section 32-11 as being responsible for ferromagnetism is not the mutual magnetic interaction between two elementary magnetic dipoles. To show this, calculate (a) the magnitude of the magnetic field a distance of \(10 \mathrm{~nm}\) away, along the dipole axis, from an atom with magnetic dipole moment \(1.5 \times 10^{-23} \mathrm{~J} / \mathrm{T}\) (cobalt), and (b) the minimum energy required to turn a second identical dipole end for end in this field. (c) By comparing the latter with the mean translational kinetic energy of \(0.040 \mathrm{eV}\), what can you conclude?

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?

The magnitude of the dipole moment associated with an atom of iron in an iron bar is \(2.1 \times 10^{-23} \mathrm{~J} / \mathrm{T}\). Assume that all the atoms in the bar, which is \(5.0 \mathrm{~cm}\) long and has a cross-sectional area of \(1.0 \mathrm{~cm}^{2}\), have their dipole moments aligned. (a) What is the dipole moment of the bar? (b) What torque must be exerted to hold this magnet perpendicular to an external field of magnitude \(1.5 \mathrm{~T}\) ? (The density of iron is \(7.9 \mathrm{~g} / \mathrm{cm}^{3} .\) )

What is the measured component of the orbital magnetic dipole moment of an electron with (a) \(m_{\ell}=1\) and (b) \(m_{\ell}=-2\) ?
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