/*! 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 149 The time period of a charged par... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 149

The time period of a charged particle undergoing a circular motion in a uniform magnetic filed is independent its (A) Speed (B) Mass (C) Charge (D) Magnetic induction

Short Answer

Expert verified
The time period of a charged particle undergoing a circular motion in a uniform magnetic field is independent of its speed. Hence, option (A) is the correct answer.

Step by step solution

01

Analyze the time period equation

Analyze the given equation for time period \(T = 2\pi\sqrt{\frac{m}{qB}}\). Clearly, this equation shows that the time period of the motion is dependent on the mass of the particle \(m\), the charge of the particle \(q\), and the magnetic field \(B\).
02

Analysing each option

(A) Speed: The speed of the particle \(v\) is not present in the time period equation, thus it is not a factor in time period of the particle's circular motion which makes this option incorrect. (B) Mass: From the equation, the time period \(T\) is directly proportional to the square root of the mass \(m\), which makes this option incorrect. (C) Charge: From the equation, the time period \(T\) is inversely proportional to the square root of the charge \(q\), which makes this option incorrect. (D) Magnetic Induction: Magnetic Induction is another term for magnetic field \(B\). The time period \(T\) is inversely proportional to the square root of the magnetic field \(B\) or magnetic induction, which makes this option incorrect as well.

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.

Time Period of Charged Particles
When a charged particle moves in a circular path within a magnetic field, you'll encounter the concept of its time period. The time period, denoted as \( T \), refers to the time it takes for the particle to complete one full circle. For charged particles, the time period can be calculated using the equation:
  • \( T = 2\pi\sqrt{\frac{m}{qB}} \)
Here, \( m \) stands for the mass of the particle, \( q \) represents the charge, and \( B \) is the magnetic field strength. You'll notice that speed doesn't appear in this equation. As a result, speed has no effect on time period. This might seem surprising, but it's a key point! The time period of a charged particle in a magnetic field depends mainly on mass, charge, and the field strength.
Uniform Magnetic Field
In physics, a uniform magnetic field is one where the magnetic field lines are parallel and equally spaced. This ensures that the magnetic field strength remains constant at every point in the area under consideration. A uniform magnetic field is crucial when discussing circular motion of charged particles because:
  • It exerts a consistent magnetic force on the particle.
  • The force modifies the path into a perfect circle.
  • The magnetic field doesn't change over time or space, making calculations more straightforward.
By maintaining a consistent magnetic field, it鈥檚 easier to predict the motion characteristics like radius and time period. For students learning about circular motion and magnetic fields, remember that uniformity simplifies the equations and outcomes.
Motion Equation Analysis
To understand the pathway that a charged particle takes, it's necessary to know how to analyze the motion equation. The equation for time period \( T = 2\pi\sqrt{\frac{m}{qB}} \) gives insight into how different factors influence the motion:
  • Mass \( (m) \): Influences the "heaviness" of the particle. A higher mass results in a longer time period.
  • Charge \( (q) \): Deals with the particle's electric property. A higher charge reduces the time period as it increases the magnetic force exerted.
  • Magnetic Field \( (B) \): Represents the magnetic environment. An increase in field strength decreases the time period.
This equation is key for entering exams or solving homework problems, as it highlights the various dependencies and allows prediction of how changes in one parameter affect the others. Simplifying the physical concept into this equation can demystify what initially seems complex.
Physics Problems
When facing physics problems, especially those involving charged particles and magnetic fields, it's vital to break down the components of the question. Analyzing the parts like in the exercise:
  • What factors are given?
  • Which parameters need predicting?
  • How best to use formulas to draw conclusions?
It's common to feel overwhelmed initially, but by understanding each term and its role, like the ones in the circular motion equation above, you can navigate these problems efficiently. Keep practicing by recognizing patterns in equations, understanding the physics behind them, and methodically working through the steps laid out in your solutions to develop confidence.

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

A bar magnet has a magnetic moment of \(2.5 \mathrm{JT}^{-1}\) and is placed in a magnetic field of \(0.2 \mathrm{~T}\). Work done in turning the magnet from parallel to antiparallel position relative to field direction is

When a long wire carrying a steady current is bent into a circular coil of one turn, the magnetic induction at its centre is \(B\). When the same wire carrying the same current is bent to form a circular coil of \(n\) turns of a smaller radius, the magnetic induction at the centre will be (A) \(B / n\) (B) \(n B\) (C) \(B / n^{2}\) (D) \(n^{2} B\)

A conductor in the form of a right angle \(A B C\) with \(A B=3 \mathrm{~cm}\) and \(B C=4 \mathrm{~cm}\) carries a current of \(10 \mathrm{~A}\). There is a uniform magnetic field of \(5 \mathrm{~T}\) perpendicular to the plane of the conductor. The force on the conductor will be (A) \(1.5 \mathrm{~N}\) (B) \(2.0 \mathrm{~N}\) (C) \(2.5 \mathrm{~N}\) (D) \(3.5 \mathrm{~N}\)

Assertion: A conducting circular disc of radius \(R\) rotates about its own axis with angular velocity \(\omega\) in uniform magnetic field \(B_{0}\) along axis of the disc then no EMF is induced in the disc. Reason: Whenever a conductor cuts across magnetic lines of flux, an EMF is induced in the conductor. (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\overline{\mathrm{C}}\) (D) D

A particle of mass \(m\) and charge \(q\) is projected from origin with initial velocity \([u \hat{i}-v \hat{j}]\). Uniform electric and magnetic field exist in the region along \(+y\) direction of magnitudes \(E\) and \(B\), respectively. Then particle will definitively return to the origin if (A) \(\frac{v B}{2 \pi E}\) is an integer (B) \(\frac{\left(u^{2}+v^{2}\right)^{1 / 2}}{[B / \pi E]}\) is an integer (C) \(\frac{v B}{\pi E}\) is an integer (D) \(\frac{v B}{3 \pi E}\) is an integer
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Scientific Method Physics

Read Explanation

Nuclear Physics

Read Explanation

Particle Model of Matter

Read Explanation

Electromagnetism

Read Explanation

Energy Physics

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.