/*! 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 15 A proton and an alpha particle e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 15

A proton and an alpha particle enter a uniform magnetic field with the same velocity. The period of rotation of the alpha particle will be (A) four times that of the proton. (B) two times that of the proton. (C) three times that of the proton. (D) same as that of the proton.

Short Answer

Expert verified
The period of rotation of the alpha particle will be four times that of the proton. To find this, we used the formula \(T = \frac{2\pi m}{qB}\) and created a ratio between the periods of rotation for both particles. After substituting the mass and charge relationships and cancelling out the common terms, we found that \(\frac{T_\alpha}{T_p} = 4\). Therefore, the correct answer is (A) four times that of the proton.

Step by step solution

01

Identify the given information and formula

We know that both the proton and the alpha particle are entering the magnetic field with the same velocity. Also, we know the period of rotation formula: \(T = \frac{2\pi m}{qB}\).
02

Write down the equation for both particles

Let's denote the period of rotation for the proton as \(T_p\) and for the alpha particle as \(T_\alpha\). Therefore, we have two equations: \(T_p = \frac{2\pi m_p}{q_pB}\) \(T_\alpha = \frac{2\pi m_\alpha}{q_\alpha B}\) Alpha particles are composed of two protons and two neutrons, so their mass is approximately 4 times the mass of a proton, and their charge is twice the charge of a proton. Let's denote mass of the proton as \(m_p\) and charge of the proton as \(q_p\), so we have: \(m_\alpha= 4m_p\), \(q_\alpha = 2q_p\)
03

Create a ratio between the periods of rotation and substitute the relevant information

To find the relation between the periods, we form the following equation: \(\frac{T_\alpha}{T_p} = \frac{\frac{2\pi m_\alpha}{q_\alpha B}}{\frac{2\pi m_p}{q_pB}}\) Now, let's substitute the mass and charge relationship: \(\frac{T_\alpha}{T_p} = \frac{\frac{2\pi(4m_p)}{2q_pB}}{\frac{2\pi m_p}{q_pB}}\) The terms \(2\pi\), \(B\), and \(q_p\) cancel out from both the numerator and the denominator: \(\frac{T_\alpha}{T_p} = \frac{4m_p}{m_p}\) Finally, we have: \(\frac{T_\alpha}{T_p} = 4\) This means the period of rotation of the alpha particle is four times that of the proton.
04

Conclusion

The correct answer is (A) four times that of the proton.

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!

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

In an oscillating \(L-C\) circuit, the maximum charge on the capacitor is \(Q\). The charge on the capacitor when the energy is stored equally between the electric and magnetic field is (A) \(\frac{Q}{2}\) (B) \(\frac{Q}{\sqrt{2}}\) (C) \(\frac{Q}{\sqrt{3}}\) (D) \(\frac{Q}{3}\)

A uniform disc of radius \(R\) having charge \(Q\) distributed uniformly all over its surface is placed on a smooth horizontal surface. A magnetic field, \(B=k x t^{2}\), where \(k\) is a constant, \(x\) is the distance (in metre) from the centre of the disc, and \(t\) is the time (in second) is switched on perpendicular to the plane of the disc. Find the torque (in \(\mathrm{N}-\mathrm{m}\) ) acting on the disc after \(15 \mathrm{~s}\) (Take \(4 k Q=1\) S.I. unit and \(R=1 \mathrm{~m}\) ).

The mutual inductance between two planar concentric rings of radii \(r_{1}\) and \(r_{2}\) (with \(r_{1} \gg r_{2}\) ) placed in air is given by (A) \(\frac{\mu_{0} \pi r_{2}^{2}}{2 r_{1}}\) (B) \(\frac{\mu_{0} \pi r_{1}^{2}}{2 r_{2}}\) (C) \(\frac{\mu_{0} \pi\left(r_{1}+r_{2}\right)^{2}}{2 r_{1}}\) (D) \(\frac{\mu_{0} \pi\left(r_{1}+r_{2}\right)^{2}}{2 r_{2}}\)

A coil having an area \(A_{0}\) is placed in a magnetic field which changes from \(B_{0}\) to \(4 B_{0}\) in time interval \(t\). The average EMF induced in the coil will be (A) \(\frac{3 A_{0} B_{0}}{t}\) (B) \(\frac{4 A_{0} B_{0}}{t}\) (C) \(\frac{3 B_{0}}{A_{0} t}\) (D) \(\frac{4 B_{0}}{A_{0} t}\)

A magnetic field \(\vec{B}=\left(\frac{B_{0} y}{a}\right) \hat{k}\) is into the paper in the \(+z\) direction. \(B_{0}\) and \(a\) are positive constants. A square loop \(E F G H\) of side \(a\), mass \(m\), and resistance \(R\), in \(x-y\) plane, starts falling under the influence of gravity. Assume \(x\)-axis is horizontal and \(y\) is vertically downward. The magnitude and direction of the induced current in the loop when its speed is \(v\) is (A) \(\frac{B_{0} a v}{R}\), anti-clockwise (B) \(\frac{B_{0} a v}{R}\), clockwise (C) \(\frac{B_{0} v}{a R}\), anti-clockwise (D) None of these
See all solutions

Recommended explanations on Physics Textbooks

Turning Points in Physics

Read Explanation

Medical Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Scientific Method Physics

Read Explanation

Electromagnetism

Read Explanation

Physical Quantities And Units

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.