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Magnetic field of earth is identical to magnetic field of a giant magnet held \(20^{\circ}\) west of geographic \(N-\mathrm{S}\) at the centre of earth. At equator, horizontal component of earth is 0.32 G. Vertical component can be calculated from the relation \(V=H \tan \delta\), where \(\delta\) is angle of dip at the place. The value of \(\delta=0^{\circ}\) at equator and \(8=90^{\circ}\) at poles. At a particle place, \(V=H\). The angle of dip is (a) 45 * (b) \(90^{\circ}\) (c) \(0^{\circ}\) (d) None of these

Short Answer

Expert verified
The angle of dip is \(45^{\circ}\) (option a).

Step by step solution

01

Identifying Known Values and Formula

We know that the angle of dip, \( \delta = 0^{\circ} \) at the equator and \( \delta = 90^{\circ} \) at the poles. We are also given that the vertical component \( V \) equals the horizontal component \( H \), so \( V = H \). The formula provided is \( V = H \tan \delta \).
02

Set Up and Solve Equation for \( \delta \)

Since \( V = H \), we substitute into the formula: \( H = H \tan \delta \). Dividing both sides by \( H \) (and assuming \( H eq 0 \)), we get \( 1 = \tan \delta \).
03

Find \( \delta \) from Trigonometric Equality

We need to find an angle \( \delta \) such that \( \tan \delta = 1 \). The angle whose tangent is 1 is \( \delta = 45^{\circ} \).

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Key Concepts

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

Horizontal Component of Earth's Magnetic Field
The Earth's magnetic field can be divided into horizontal and vertical components. The horizontal component, often denoted as \( H \), represents the magnetic field's strength parallel to the Earth's surface. As such, it is pivotal in navigational practices, particularly in areas close to the equator where the horizontal component determines compass readings.
At the equator, the horizontal component plays a dominant role. For instance, in the exercise, the given horizontal component at the equator is 0.32 G (Gauss). This is crucial because it provides an indication of how much of the Earth's magnetic force is exerted in a horizontal direction, affecting how compasses respond in this region.

Some key things to consider about the horizontal component:
  • The strength of the horizontal component can vary depending on the geographic location, especially from equator to poles.
  • The horizontal component is influenced by the angle of the Earth's magnetic inclination or dip.
  • Navigation and satellite guidance systems often rely on understanding the variations in the horizontal component for accuracy.
Understanding \( H \) is essential for interpreting magnetic field-related phenomena and applications.
Angle of Dip
The angle of dip, also known as the magnetic inclination, is an angle that defines the tilt of the Earth's magnetic field lines concerning the horizontal plane. At any given location, it tells us how steeply the magnetic lines of force enter or leave the Earth’s surface and is denoted by the symbol \( \delta \).
This angle is critical because it can vary significantly around the planet. For example, at the equator, this angle is \( 0^{\circ} \) because the magnetic field lines are perfectly parallel to the surface. Meanwhile, at the poles, the angle is \( 90^{\circ} \), indicating that the lines are perpendicular.

In the provided exercise, calculating the angle of dip is required when the vertical and horizontal components are equal. The mathematical expression \( V = H \tan \delta \) is used, leading to a simplified equation \( 1 = \tan \delta \), suggesting \( \delta = 45^{\circ} \).
Some important points about the angle of dip include:
  • A crucial factor in determining the behavior of the magnetic field at different latitudes.
  • It influences the vertical and horizontal components, crucial for calculating the Earth's magnetic field at a given location.
  • Magnetometers often measure this angle to assist in accuracy for geological and navigational purposes.
Understanding the angle of dip provides vital insights into both the orientation and strength of the Earth’s magnetic field.
Vertical Component of Earth's Magnetic Field
The vertical component of the Earth's magnetic field, denoted by \( V \), signifies the magnetic field's strength perpendicular to the Earth’s surface. It plays a particularly crucial role near the poles, where the magnetic field lines enter or exit vertically.
In the given problem, the vertical component can be calculated using the equation \( V = H \tan \delta \). This relationship shows how the vertical component depends on both the horizontal component \( H \) and the angle of dip \( \delta \).

For example, knowing that at certain locations \( V = H \), it allows us to determine specific characteristics of the Earth's magnetic field. In this case, it is discovered that \( \delta = 45^{\circ} \) when \( V = H \), as the tangent of \( 45^{\circ} \) is equal to 1, satisfying the equation.
The vertical component impacts many fields and applications:
  • Essential for understanding the directional properties of the Earth's magnetic field.
  • Influential in various navigational, geological, and atmospheric studies.
  • Influences compass readings and other magnetic instruments when the angle of dip is significant.
  • Comprehending \( V \) is key to grasping the full dynamics of the Earth’s magnetic field and its varied effects across the globe.

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    Most popular questions from this chapter

    In an experiment with vibration magnetometer, the value of \(4 \pi^{2} I / T^{2}\) for a short bar magnet is observed as \(36 \times 10^{-4}\). In the experiment with deflection magnetometer with the same magnet, the value of \(4 \pi d^{3} / 2 \mu_{0}\) is observed as \(10^{8} / 36\). The magnetic moment of the magnet used is (a) \(50 \mathrm{~A}-\mathrm{m}\) (b) \(100 \mathrm{~A}-\mathrm{m}\) (c) \(200 \mathrm{~A}-\mathrm{m}\) (d) \(1000 \mathrm{~A}-\mathrm{m}\)

    The magnetic field of the earth can be modelled by that of a point dipole placed at the centre of the earth. The dipole axis makes an angle of \(11,3^{\circ}\) with the axis of the earth. At Mumbai, declination is nearly zero.Then. INCFRT Exemplar] (a) the declination varies between \(11.3^{*} \mathrm{~W}\) to \(11,3^{-} \mathrm{E}\) (b) the least declination is \(0^{*}\) (c) the plane defined by dipole axis and Earth axis passes through Grecmwich (d) declination averaged over Earth must be always negative

    The magnetic needle of a tangent galvanometer is deflected at an angle \(30^{\circ}\) due to a magnet. The horizontal component of earth's magnetic field \(0.34 \times 10^{-4} \mathrm{~T}\) is along the plane of the coil. The magnetic intensity is (a) \(1.96 \times 10^{-4} \mathrm{~T}\) (b) \(1.96 \times 10^{4} \mathrm{~T}\) (c) \(1.96 \times 10^{-5} \mathrm{~T}\) (d) \(1.96 \times 10^{5} \mathrm{~T}\)

    A long magnet is placed vertically with its \(S\)-pole resting on the table. A neutral point is obtained \(10 \mathrm{em}\) from the pole due geographie north of it. If \(H=3.2 \times 10^{-5} \mathrm{~T}\), then the pole strength of magnet is (a) \(8 \mathrm{ab}-\mathrm{A}-\mathrm{cm}^{-1}\) (b) \(16 a b-\mathrm{A}-\mathrm{cm}^{-1}\) (c) \(32 \mathrm{ab}-\mathrm{A}-\mathrm{cm}^{-1}\) (d) \(64 \mathrm{ab}-\mathrm{A}-\mathrm{cm}^{-1}\)

    A uniform magnetic needle is suspended from its centre by a thread. Its upper end is now loaded with a mass of \(50 \mathrm{mg}\), and the needle becomes horizontal. If the strength of each pole is \(98.1\) ab-amp-cm and \(g=981 \mathrm{cms}^{-2}\), then the vertical component of earth's magnetic induction is (a) \(0.50 \mathrm{G}\) (b) \(0.25 \mathrm{G}\) (c) \(0.005 \mathrm{G}\) (d) \(0.05 \mathrm{G}\)

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