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Chapter 20: Problem 15

The earth's magnetic field at a certain place has a horizontal component of \(0.3 \mathrm{G}\) and total strength \(0.5 \mathrm{G}\). Find angle of dip in \(\tan ^{-1}\). (a) \(\delta=\tan ^{-1} \frac{4}{3}\) (b) \(\delta=\tan ^{-1} \frac{3}{4}\) (c) \(\vec{b}=\tan ^{-1} \frac{5}{3}\) (d) \(\delta=\tan ^{-1} \frac{3}{5}\)

Short Answer

Expert verified
The correct answer is (a) \( \delta = \tan^{-1}(\frac{4}{3}) \).

Step by step solution

01

Understanding the Problem

We need to find the angle of dip (also known as the magnetic inclination) given the horizontal component of Earth's magnetic field and the total magnetic field strength. The angle of dip is denoted by \( \delta \) and is given by the equation \( \tan(\delta) = \frac{B_v}{B_h} \), where \( B_v \) is the vertical component and \( B_h \) is the horizontal component. We have \( B_h = 0.3 \; \mathrm{G} \) and the total magnetic field \( B = 0.5 \; \mathrm{G} \).
02

Express the Vertical Component

Using the Pythagorean theorem for vector components, we have the equation: \( B = \sqrt{B_h^2 + B_v^2} \). Substituting the known values, \(0.5 = \sqrt{0.3^2 + B_v^2}\). We need to solve this to find \( B_v \).
03

Solve for Vertical Component

Squaring both sides of the equation yields \( 0.25 = 0.09 + B_v^2 \). Solving for \( B_v^2 \) gives \( B_v^2 = 0.25 - 0.09 = 0.16 \). Thus, \( B_v = \sqrt{0.16} = 0.4 \; \mathrm{G} \).
04

Find the Angle of Dip

We now use the equation for the angle of dip: \( \tan(\delta) = \frac{B_v}{B_h} = \frac{0.4}{0.3} = \frac{4}{3} \). Thus, the angle of dip \( \delta = \tan^{-1}\left(\frac{4}{3}\right) \).
05

Select the Correct Answer

Review the options provided: (a) \( \delta=\tan ^{-1} \frac{4}{3} \), (b) \( \delta=\tan ^{-1} \frac{3}{4} \), (c) \( \vec{b}=\tan ^{-1} \frac{5}{3} \), (d) \( \delta=\tan ^{-1} \frac{3}{5} \). The correct answer is option (a): \( \delta=\tan ^{-1} \frac{4}{3} \).

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Key Concepts

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

Magnetic Field Components
Understanding the magnetic field components is fundamental to solving the angle of dip problem. The Earth's magnetic field at any location can be split into two main components: horizontal and vertical. The horizontal component (\(B_h\)) is the part of the magnetic field that runs parallel to the surface of the Earth.
The vertical component (\(B_v\)) is perpendicular to the Earth's surface.

Both components are crucial in determining the overall strength and direction of the magnetic field at a point. Knowing these components helps in calculating the magnetic inclination (angle of dip), where the vertical component plays a significant role. In our exercise, the horizontal component was given as \(0.3 \; \mathrm{G}\), and through calculations, we found the vertical component to be \(0.4 \; \mathrm{G}\), leading to the determination of the angle of dip.
Pythagorean Theorem
The Pythagorean theorem is a key mathematical principle used to find unknown components when dealing with right-angled triangles. It states that in a right-angled triangle, the sum of the squares of the two perpendicular sides is equal to the square of the hypotenuse.
In the context of magnetic field components, the total magnetic field strength (\(B\)) acts as the hypotenuse, while the horizontal (\(B_h\)) and vertical (\(B_v\)) components are the perpendicular sides of a right-angled triangle.

In our exercise, we applied the Pythagorean theorem to find the vertical component from the total field: \[B = \sqrt{B_h^2 + B_v^2}\]Substituting the given values, we solved \(0.25 = 0.09 + B_v^2\) to find \(B_v = 0.4 \; \mathrm{G}\), which was pivotal in discovering the angle of dip.
Magnetic Inclination
Magnetic inclination, or the angle of dip, is the angle that the Earth's magnetic field makes with the horizontal plane at a specific location. It gives valuable information about the field's orientation.
The angle of dip helps in understanding how the magnetic field 'plunges' into the Earth. Regions near the magnetic poles have a steep dip, while those near the equator have a shallower one.

To find the angle of dip, you use the relation \(\tan(\delta) = \frac{B_v}{B_h}\), where \(B_v\) is the vertical component and \(B_h\) is the horizontal component. For instance, in the exercise given, the use of the formula allowed us to conclude that \(\delta = \tan^{-1}(\frac{4}{3})\), illustrating the dip angle at the specific location.
Trigonometric Ratios
Trigonometric ratios are mathematical functions that establish relationships between the sides of a right-angled triangle and its angles. The primary trigonometric ratios are sine, cosine, and tangent.
In our exercise, the tangent ratio (\(\tan\)) was used to find the angle of dip (magnetic inclination).

The tangent of an angle in a right triangle is defined as the ratio of the opposite side to the adjacent side. Thus, for magnetic inclination, \(\tan(\delta) = \frac{B_v}{B_h}\). This allows us to calculate the angle of dip when the two components are known.
This simple yet powerful mathematical tool gave us \(\delta = \tan^{-1}(\frac{4}{3})\), providing the precise angle at which the magnetic field dips into the Earth.

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Most popular questions from this chapter

A magnetic dipole is placed at right angles to the direction of lines of force of magnetic induction \(B\). If it is rotated through an angle of \(180^{\circ}\), then the work done is (a) \(M B\) (b) \(2 \mathrm{MB}\) (c) \(-2 \mathrm{MB}\) (d) zero

A bar magnet has a magnetic moment equal to \(5 \times 10^{-5} \mathrm{~Wb}-\mathrm{m} .\) It is suspended in a magnetic field which has a magnetic induetion \(B\) equal to \(8 \pi \times 10^{-4} \mathrm{~T}\). The magnet vibrates with a period of vibration equal to 15 s. The moment of inertia of magnet is (a) \(4.54 \times 10^{4} \mathrm{~kg}-\mathrm{m}^{2}\) (b) \(4.54 \times 10^{-5} \mathrm{~kg}-\mathrm{m}^{2}\) (c) \(4.54 \times 10^{-4} \mathrm{~kg}-\mathrm{m}^{2}\) (d) \(4.54 \times 10^{5} \mathrm{~kg}-\mathrm{m}^{2}\)

Which of the following characteristie is not associated with a ferromagnetic material? [Kerala CET 2008] (a) It is strongly attracted by a magnet (b) It tends to move from a region of strong magnetic ficld to a region of weak magnetic ficld (c) Its origin is the spin of clectrons (d) Above the curie temperature, it exhibits paramagnetic properties

The relative permeability of a substance \(X\) is slightly less than unity and that of aubstance \(Y\) is slightly more than unity, then (a) \(X\) is paramagnetic and \(Y\) is ferromagnetic (b) \(X\) is diamagnetic and \(Y\) is ferromagnetic (c) \(X\) and \(Y\) both are paramagnetic (d) \(X\) is diamagnetic and \(Y\) is paramagnetic

In a certain place, the horizontal component of magnetic field is \(\frac{1}{\sqrt{3}}\) times the vertical component. The angle of dip at this place is (a) zero (b) \(\pi / 3\) (c) \(\pi / 2\) (d) \(\pi / 6\)
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