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The true value of angle of dip at a place is \(60^{\circ}\), the apparent dip in a plane inclined at an angle of \(30^{\circ}\) with magnetic meridian is. (a) \(\tan ^{-1}(1 / 2)\) (b) \(\tan ^{-1} 2\) (c) \(\tan ^{-1}(2 / 3)\) (d) None of these

Short Answer

Expert verified
The apparent dip in a plane inclined at an angle of \(30^{\circ}\) with the magnetic meridian is \(\tan^{-1}(\frac{\sqrt{3}}{2})\).

Step by step solution

01

Write down the given values

We have the following values given: True dip (δ) = \(60^{\circ}\) Inclination of the plane (β) = \(30^{\circ}\)
02

Substitute the given values in the formula

We have the formula: \( \tan\alpha = \tan\delta \sin\beta \). Substitute the given values of true dip and inclination of the plane: \(\tan\alpha = \tan(60^{\circ}) \sin(30^{\circ})\)
03

Calculate the value of tanα

Now, calculate the value of \(\tan\alpha\): \( \tan\alpha = \tan(60^{\circ}) \sin(30^{\circ}) = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2} \)
04

Find the angle α

To find the angle α, we need to find the inverse tangent of the obtained value: \(\alpha = \tan^{-1}(\frac{\sqrt{3}}{2})\)
05

Choose the correct option

From the given choices, option (c) is \(\tan ^{-1}(2 / 3)\), which is not equal to \(\tan^{-1}(\frac{\sqrt{3}}{2})\). Therefore, the correct answer is option (d) None of these. The apparent dip in a plane inclined at an angle of \(30^{\circ}\) with the magnetic meridian is \(\tan^{-1}(\frac{\sqrt{3}}{2})\).

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Key Concepts

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

Angle of Dip
In the world of magnetism, the angle of dip, sometimes referred to as magnetic inclination, is the angle made by the Earth's magnetic field lines with the horizontal plane at any given point on the Earth's surface. It's a crucial parameter for understanding how Earth's magnetic field interacts with objects on the surface.

The angle of dip varies from place to place due to the structure of the Earth's magnetic field. At the magnetic poles, the angle of dip is almost 90 degrees because the magnetic field lines are nearly vertical. At the magnetic equator, the field lines run parallel to the surface, resulting in a dip of zero.

For practical purposes, the angle of dip helps in navigation and understanding the magnetic environment. Aviation and maritime industries, for example, rely on dip readings for accurate compass calibration.
Apparent Dip
Apparent dip is a concept used to describe what happens when a plane is tilted with respect to the magnetic meridian. Imagine you're in a place where the true dip is known, but the surface you're measuring it on isn't flat or aligned with the magnetic meridian.

When the plane is inclined, the observed angle, called the apparent dip, will differ from the true dip. This occurs because the tilt of the plane changes the interaction with the magnetic field. Calculating apparent dip requires adjustments using trigonometric relationships. The formula used is:
  • \( \tan\alpha = \tan\delta \sin\beta \)
where
  • \( \alpha \) is the apparent dip,
  • \( \delta \) is the true dip,
  • \( \beta \) is the angle between the plane and the magnetic meridian.

Understanding apparent dip is essential for geological studies and compass readings when the local terrains are uneven or structures are inclined.
Magnetic Meridian
The term 'magnetic meridian' refers to an imaginary line around Earth, which connects the magnetic north and south poles. Think of it as an invisible vertical plane through which Earth's magnetic field lines travel.

A compass needle aligns itself with the magnetic meridian because it aligns with the Earth's magnetic field. This makes the concept crucial for navigation. Aligning measurements or devices to the magnetic meridian ensures that the magnetic field lines are accurately followed.

In practical applications, magnetic meridians are used to ensure instruments and measurements are correctly orientated. This is essential not only for navigation but also for geological and geophysical surveys, where understanding Earth's magnetic influence is paramount.

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Most popular questions from this chapter

Two similar coils are kept mutually perpendicular such that their centers co- inside. At the centre, find the ratio of the mag. field due to one coil and the resultant magnetic field by both coils, if the same current is flown. (a) \(1: \sqrt{2}\) (b) \(1: 2\) (c) \(2: 1\) (d) \(\sqrt{3}: 1\)

A proton is projected with a speed of \(2 \times 10^{6}(\mathrm{~m} / \mathrm{s})\) at an angle of \(60^{\circ}\) to the \(\mathrm{X}\) -axis. If a uniform mag. field of \(0.104\) tesla is applied along \(\mathrm{Y}\) -axis, the path of proton is (a) A circle of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period \(\pi \times 10^{-7} \mathrm{sec}\) (b) A circle of \(\mathrm{r}=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{sec}\) (c) A helix of \(r=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{sec}\) (d) A helix of \(\mathrm{r}=0.2 \mathrm{~m}\) and time period \(4 \pi \times 10^{-7} \mathrm{sec}\)

A particle of mass \(\mathrm{m}\) and charge q moves with a constant velocity v along the positive \(\mathrm{x}\) -direction. It enters a region containing a uniform magnetic field B directed along the negative \(z\) -direction, extending from \(x=a\) to \(x=b\). The minimum value of required so that the particle can just enter the region \(\mathrm{x}>\mathrm{b}\) is (a) \([(\mathrm{q} \mathrm{bB}) / \mathrm{m}]\) (b) \(q(b-a)(B / m)\) (c) \([(\mathrm{qaB}) / \mathrm{m}]\) (d) \(q(b+a)(B / 2 m)\)

In each of the following questions, Match column-I and column-II and select the correct match out of the four given choices.(B) Ammeter (Q) Moderate resistance (C) Voltmeter (R) High, Low or moderate resistance (D) Avometer (S) High resistance (a) \(\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}\) (b) \(\mathrm{A} \rightarrow \mathrm{P} ; \mathrm{B} \rightarrow \mathrm{Q} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}\) (c) \(\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{R} ; \mathrm{D} \rightarrow \mathrm{S}\) (d) \(\mathrm{A} \rightarrow \mathrm{Q} ; \mathrm{B} \rightarrow \mathrm{P} ; \mathrm{C} \rightarrow \mathrm{S} ; \mathrm{D} \rightarrow \mathrm{R}\)

Two concentric co-planar circular Loops of radii \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) carry currents of respectively \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\) in opposite directions. The magnetic induction at the centre of the Loops is half that due to \(\mathrm{I}_{1}\) alone at the centre. If \(\mathrm{r}_{2}=2 \mathrm{r}_{1}\) the value of \(\left(\mathrm{I}_{2} / \mathrm{I}_{1}\right)\) is (a) 2 (b) \(1 / 2\) (c) \(1 / 4\) (d) 1

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