Log out Go to App
Go to App

Learning Materials

Features

Discover

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) \(45^{\circ}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The unit of ele. current "AMPERE" is the current which when flowing through each of two parallel wires spaced 1 meter apart in vacuum and of infinite length will give rise to a force between them equal to $\mathrm{N} / \mathrm{m}$ (a) 1 (b) \(2 \times 10^{-7}\) (c) \(1 \times 10^{-2}\) (d) \(4 \pi \times 10^{-7}\)

The effective length of a magnet is \(31.4 \mathrm{~cm}\) and its pole strength is \(0.5 \mathrm{~A} \mathrm{~m}\). The magnetic moment, if it is bent in the form of a semicircle will be Amp.m \(^{2}\) (a) \(0.1\) (b) \(0.01\) (c) \(0.2\) (d) \(1.2\)

A small bar magnet has a magnetic moment $1.2 \mathrm{~A} \cdot \mathrm{m}^{2}\(. The magnetic field at a distance \)0.1 \mathrm{~m}$ on its axis will be tesla. (a) \(1.2 \times 10^{-4}\) (b) \(2.4 \times 10^{-4}\) (c) \(2.4 \times 10^{4}\) (d) \(1.2 \times 10^{4}\)

A magnetic field existing in a region is given by $\mathrm{B}^{-}=\mathrm{B}_{0}[1+(\mathrm{x} / \ell)] \mathrm{k} \wedge . \mathrm{A}\( square loop of side \)\ell$ and carrying current I is placed with edges (sides) parallel to \(\mathrm{X}-\mathrm{Y}\) axis. The magnitude of the net magnetic force experienced by the Loop is (a) \(2 \mathrm{~B}_{0} \overline{\mathrm{I} \ell}\) (b) \((1 / 2) B_{0} I \ell\) (c) \(\mathrm{B}_{\circ} \mathrm{I} \ell\) (d) BI\ell

Needles \(\mathrm{N}_{1}, \mathrm{~N}_{2}\) and \(\mathrm{N}_{3}\) are made of a ferromagnetic, a paramagnetic and a diamagnetic substance respectively. A magnet when brought close to them will (a) Attract \(\mathrm{N}_{1}\) strongly, \(\mathrm{N}_{2}\) weakly and repel \(\mathrm{N}_{3}\) weakly (b) Attract \(\mathrm{N}_{1}\) strongly, but repel \(\mathrm{N}_{2}\) and \(\mathrm{N}_{3}\) weakly (c) Attract all three of them (d) Attract \(\mathrm{N}_{1}\) and \(\mathrm{N}_{2}\) strongly but repel \(\mathrm{N}_{2}\)
See all solutions

Recommended explanations on English Textbooks

English Grammar

Read Explanation

Lexis and Semantics

Read Explanation

Linguistic Terms

Read Explanation

Essay Plan

Read Explanation

Morphology

Read Explanation

Text Comparison

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free