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Chapter 13: Problem 1976

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be \(2 \mathrm{sec}\). The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be \(2 \mathrm{sec}\). Then the angle of dip is (a) \(0^{\circ}\) (b) \(30^{\circ}\) (c) \(45^{\circ}\) (d) \(90^{\circ}\)

Short Answer

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

Step by step solution

01

Formula for time period in vertical plane

In the vertical plane (perpendicular to magnetic meridian), the time period formula of vibration is given by the following relation: \(T_v = 2\pi\sqrt{\frac{I}{mH_v}}\) Where \(T_v\) is the time period in the vertical plane, \(I\) is the moment of inertia of the dip needle, \(m\) is the needle's magnetic moment, and \(H_v\) is the vertical component of the earth's magnetic field.
02

Formula for time period in horizontal plane

In the horizontal plane, the time period formula of vibration is given by the following relation: \(T_h = 2\pi\sqrt{\frac{I}{mH_h}}\) Where \(T_h\) is the time period in the horizontal plane and \(H_h\) is the horizontal component of the earth's magnetic field.
03

Given values of time periods

We are given that the time periods in both planes are the same, i.e., \(T_v = T_h = 2\) seconds. This means that: \(2\pi\sqrt{\frac{I}{mH_v}} = 2\pi\sqrt{\frac{I}{mH_h}}\)
04

Simplifying the equation

By simplifying the equation, we get: \(\frac{1}{\sqrt{H_v}} = \frac{1}{\sqrt{H_h}}\) Squaring both sides, we obtain: \(\frac{1}{H_v} = \frac{1}{H_h}\)
05

Using the angle of dip in magnetic field components

The angle of dip (\(\delta\)) is related to the magnetic field components as follows: \(H_v = H\sin\delta\) \(H_h = H\cos\delta\) Where H is the total magnetic field at that point.
06

Substituting values into the equation

By substituting the expressions for \(H_v\) and \(H_h\) in terms of the angle of dip, we get: \(\frac{1}{H\sin\delta} = \frac{1}{H\cos\delta}\)
07

Simplifying and solving for the angle of dip

By simplifying and solving for the angle of dip (\(\delta\)), we obtain: \(\sin\delta = \cos\delta\) Since \(\sin\delta = \cos\delta\), we can rewrite the equation as \(\sin\delta = \sin\left(\frac{\pi}{2} - \delta\right)\). By using the property \(\sin\alpha=\sin\beta\), we get: \(\delta = \frac{\pi}{2} - \delta\), this results in \(2\delta = \frac{\pi}{2}\), which implies: \(\delta = \frac{\pi}{4}\) Thus, the angle of dip (\(\delta\)) is \(\frac{\pi}{4}\) radians, or \(45^{\circ}\). Therefore, the correct answer is: (c) \(45^{\circ}\)

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Key Concepts

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

Magnetic Meridian
Imagine you're using a compass to find direction. The magnetic meridian is like an invisible line that the compass needle aligns with. It's the vertical plane that passes through the north and south magnetic poles of the Earth.
This line is where a freely suspended magnet will naturally align itself. When a dip needle, a special compass, vibrates perpendicularly to this plane, it helps measure important aspects of Earth's magnetic field.
Understanding the magnetic meridian is vital for interpreting how different components of the magnetic field behave. It influences how the magnetic field's components are measured and categorized.
Vertical and Horizontal Components of Magnetic Field
The Earth's magnetic field can be broken down into two components: vertical (\(H_v\)) and horizontal (\(H_h\)) components. These components help us understand how the magnetic field interacts with objects like a dip needle.
In a vertical plane, the vertical component is powerful and affects how the needle vibrates. Similarly, in the horizontal plane, it's the horizontal component that plays a big role.
  • The vertical component (\(H_v\)) is associated with forces acting directly downward.
  • The horizontal component (\(H_h\)) runs parallel to the Earth's surface.
By comparing these components, we can calculate properties like the angle of dip, which tells us how steeply the magnetic field lines enter the Earth.
Time Period of Vibration
The time period of vibration is a measure of how long it takes for a vibrating object, like a dip needle, to complete one full cycle of motion. This is important in magnetic field studies for understanding the needle's oscillation frequency.
In both the vertical and horizontal planes, we use the time period formula to measure these vibrations:
  • Vertical plane: \(T_v = 2\pi\sqrt{\frac{I}{mH_v}}\)
  • Horizontal plane: \(T_h = 2\pi\sqrt{\frac{I}{mH_h}}\)
The comparison of these time periods helps to decode the relationship between the needle's movement and the magnetic field. A key insight from the exercise reveals equal time periods in both planes, indicating specific values for the angle of dip.
Moment of Inertia
Moment of inertia (\(I\)) is a concept from physics that defines how difficult it is to change the rotational speed of an object. Think of it like the rotational equivalent of mass; it dictates the resistance to change in motion.
When dealing with a dip needle, the moment of inertia plays a crucial role in its vibration characteristics. The formula incorporating moment of inertia shows how it influences the time period of vibration:
  • The higher the moment of inertia, the slower the needle vibrates.
  • It depends on the distribution of the needle's mass relative to the axis of rotation.
The moment of inertia is an essential factor in these calculations, helping us connect the physical properties of the needle with its motion in the magnetic field.

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Most popular questions from this chapter

Two straight long conductors \(\mathrm{AOB}\) and \(\mathrm{COD}\) are perpendicular to each other and carry currents \(\mathrm{I}_{1}\) and \(\mathrm{I}_{2}\). The magnitude of the mag. field at a point " \(\mathrm{P}^{n}\) at a distance " \(\mathrm{a}^{\prime \prime}\) from the point "O" in a direction perpendicular to the plane \(\mathrm{ABCD}\) is (a) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}+I_{2}\right)\) (b) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(I_{1}-I_{2}\right)\) (c) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left(\mathrm{I}_{1}^{2}+\mathrm{I}_{2}^{2}\right)^{1 / 2}\) (d) \(\left[\left(\mu_{0}\right) /(2 \pi a)\right]\left[\left(I_{1} I_{2}\right) /\left(I_{1}-I_{2}\right)\right]\)

At a distance of \(10 \mathrm{~cm}\) from a long straight wire carrying current, the magnetic field is \(4 \times 10^{-2}\). At the distance of \(40 \mathrm{~cm}\), the magnetic field will be Tesla. (a) \(1 \times 10^{-2}\) (b) \(2 \times \overline{10^{-2}}\) (c) \(8 \times 10^{-2}\) (d) \(16 \times 10^{-2}\)

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An electron having mass \(9 \times 10^{-31} \mathrm{~kg}\), charge \(1.6 \times 10^{-19} \mathrm{C}\) and moving with a velocity of \(10^{6} \mathrm{~m} / \mathrm{s}\) enters a region where mag. field exists. If it describes a circle of radius \(0.10 \mathrm{~m}\), the intensity of magnetic field must be Tesla (a) \(1.8 \times 10^{-4}\) (b) \(5.6 \times \overline{10^{-5}}\) (c) \(14.4 \times 10^{-5}\) (d) \(1.3 \times 10^{-6}\)
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