/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A magnet performs 10 oscillation... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 33

A magnet performs 10 oscillations per minute in a horizontal plane at a place where the angle of dip is \(45^{\circ}\) and the total intensity is \(0.707\) CGS units. The number of oscillations per minute at a place where dip angle is \(60^{\circ}\) and total intensity is \(0.5\) CGS units will be (a) 5 (b) 7 (c) 9 (d) 11

Short Answer

Expert verified
7 oscillations per minute.

Step by step solution

01

Understanding the Problem and Given Data

We have a magnet oscillating in a horizontal plane performing 10 oscillations per minute. The angle of dip and total intensity at the first location are given as \(45^{\circ}\) and 0.707 CGS units, respectively. Our goal is to find out the number of oscillations per minute at a location where the angle of dip is \(60^{\circ}\) and the total intensity is 0.5 CGS units.
02

Relating Oscillation Frequency to Intensity

The frequency of oscillation is related to the horizontal component of the Earth's magnetic field \(B_H\) which can be found using the formula \(B_H = B_T \cos(\theta)\), where \(B_T\) is the total magnetic field intensity and \(\theta\) is the angle of dip. The frequency \(f\) of oscillations is proportional to \(\sqrt{B_H}\).
03

Calculating Horizontal Component at First Location

At the first location, the angle of dip \(\theta = 45^{\circ}\) and total intensity \(B_{T1} = 0.707\) CGS units. Thus, \(B_{H1} = 0.707 \times \cos(45^{\circ}) = 0.707 \times \frac{1}{\sqrt{2}} = 0.5\).
04

Calculating Horizontal Component at Second Location

At the second location, the angle of dip \(\theta = 60^{\circ}\) and total intensity \(B_{T2} = 0.5\) CGS units. Thus, \(B_{H2} = 0.5 \times \cos(60^{\circ}) = 0.5 \times 0.5 = 0.25\).
05

Comparing Oscillation Frequencies at Both Locations

The frequency of oscillations is proportional to \(\sqrt{B_H}\). Therefore, \(\frac{f_2}{f_1} = \sqrt{\frac{B_{H2}}{B_{H1}}} = \sqrt{\frac{0.25}{0.5}} = \sqrt{0.5} \approx 0.7071\).
06

Finding the Oscillation at the New Location

The number of oscillations (\(f_2\)) at the second location can be calculated as \(f_2 = f_1 \times \sqrt{0.5}\). With \(f_1 = 10\) oscillations per minute, \(f_2 \approx 10 \times 0.7071 \approx 7.071\). Thus, the closest integer value is 7.

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Angle of Dip
The angle of dip, also known as magnetic inclination, is the angle made by the Earth's magnetic field lines with the horizontal plane. Picture the angle formed between a compass needle that tilts freely in three dimensions and the horizontal ground surface.
  • If the angle of dip is 0掳, the magnetic field is entirely horizontal.
  • At 90掳, it is completely vertical.
  • At angles in between, such as 45掳 or 60掳, the field's orientation is diagonal.
The variation in the angle of dip at different locations is crucial for determining the horizontal component of the magnetic field, which in turn affects the oscillation frequency of a compass needle or magnet. In our problem, we see how changing the angle of dip from 45掳 to 60掳 influences the magnet's oscillation.
Earth's Magnetic Field
Earth's magnetic field is like a gigantic magnet surrounding the planet. It has both horizontal and vertical components, which vary depending on geographic location. The horizontal component is crucial when discussing magnetic oscillations because it influences a magnet's oscillation frequency.
Understanding the Earth's magnetic field helps quantify how a magnet slows down or speeds up its oscillations based on location. Scientists categorize the magnetic intensity using units like CGS, helping create formulas that relate the dip angle and total magnetic intensity to the horizontal component. In practical problems, factor locations, such as those with different dip angles and intensities, to explain their impact on physical phenomena like oscillations.
Oscillation Frequency
Oscillation frequency refers to how many times an object, like a magnet, completes a cycle in a specific time frame. When considering Earth's magnetic field, the frequency of oscillation depends primarily on the horizontal component of the magnetic field. This is because the field influences how much torque acts on the magnet.
To find out how the frequency changes, consider the relationship:
  • The frequency is proportional to the square root of the horizontal component of the magnetic field.
  • This means if the horizontal component is halved, the frequency will be multiplied by the square root of half, or approximately 0.7071.
In our problem, the principle helps calculate that changing locations will reduce oscillations from 10 to about 7 per minute.
Magnetic Field Intensity
Magnetic field intensity is quantified as the strength and direction of Earth's magnetic influence at a particular point. In our problem, we use CGS units, a traditional unit of measure derived from centimeters, grams, and seconds. This intensity is significant in determining the horizontal component of the field, using the formula:\[ B_H = B_T \cos(\theta) \]Here, \(B_H\) is the horizontal magnetic intensity, \(B_T\) the total magnetic intensity, and \(\theta\) the angle of dip.
This formula allows determination of the effective horizontal intensity that affects the oscillation of magnets. By substituting different \(B_T\) and \(\theta\), you can observe how the oscillation frequency changes, emphasizing how interconnected these concepts are.

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

Assertion-Reason type. Each of these contains two Statements: Statement 1 (Assertion), Statement II (Reason). Each of these questions also has four alternative choice, only one of which is correct. You have to select the correct choices from the codes (a), (b), (c) and (d) given below (a) If both Assertion and Reason are true and the Reason is correct explanation of the Assertion. (b) If both Assertion and Reason are true but Reason is not correct explanation of the Assertion. (c) If Assertion is true but Reason is false. (d) If Assertion is false but the Reason is true. Assertion An iron bar magnet is heated to \(1000^{\circ} \mathrm{C}\) and then cooled in a magnetic field in free space. It does not retain its magnetism. Reason Only ferro-magnets can show hysteresis.

A magnet. \(20 \mathrm{~cm}\) long with its poles concentrated at its ends is placed vertically with its north pole on the table. At a point due \(20 \mathrm{~cm}\) south (magnetic) of the pole, a neutral point is obtained. If \(H=0.3 \mathrm{G}\), then the pole strength of the magnet is approximately (a) \(185 \mathrm{ab}-\mathrm{amp}-\mathrm{cm}\) (b) \(185 \mathrm{amp}-\mathrm{m}\) (c) \(18.5 \mathrm{ab}-\mathrm{amp-cm}\) (d) \(18.5 \mathrm{amp-m}\)

Two magnets held together in earth's magnetic field with same polarity together make 12 vib-min \(^{-1}\) and when opposite poles together make 4 vib- min \(^{-1}\). The ratio of magnetie moments is (a) \(9: 1\) (b) \(1: 3\) (c) \(1: 9\) (d) \(5: 4\)

The time of vibration of a dip needle vibrating in the vertical plane is 3s. When magnetic needle is made to vibrate in the horizontal plane, the time of vibration is \(3 \sqrt{2}\) s. Then the angle of dip is (a) \(30^{\circ}\) (b) \(45^{\circ}\) (c) 60 (d) \(90^{\circ}\)

A horizontal straight wire \(20 \mathrm{~m}\) long extending from east to west is falling with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\), at right angles to the horizontal component of the earth's magnetic field0.30 \(\times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\). The instantaneous value of the emf induced in the wire will be [AIFE 2011] (a) \(6.0 \mathrm{mV}\) (b) \(3 \mathrm{mV}\) (c) \(4.5 \mathrm{mV}\) (d) \(1.5 \mathrm{mV}\)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Linear Momentum

Read Explanation

Fluids

Read Explanation

Medical Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Electrostatics

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.