/*! 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 146 A proton is fired with a speed o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 146

A proton is fired with a speed of \(5 \times 10^{7} \mathrm{~m} / \mathrm{s}\) at an angle of \(30^{\circ}\) to a magnetic field \(\vec{B}=0.40 \hat{i} \mathrm{~T}\). The pitch of the proton will be (in \(\mathrm{cm}\) )

Short Answer

Expert verified
The pitch of the proton's helical path is approximately \(1.05 \times 10^7 \mathrm{~cm}\).

Step by step solution

01

Resolve the velocity of the proton into components

We have a velocity of \(5 \times 10^7 \mathrm{~m/s}\) at an angle of \(30^{\circ}\) to the magnetic field. Let's find its components as follows: 1. Component parallel to the magnetic field, \(v_{\parallel}\): \[v_{\parallel} = v \cos{30^{\circ}}\] 2. Component perpendicular to the magnetic field, \(v_{\perp}\): \[v_{\perp} = v \sin{30^{\circ}}\] where \(v\) is the initial velocity of the proton.
02

Calculate the cyclotron frequency and the period

The cyclotron frequency, \(f_c\), is given by: \[f_c = \frac{qB}{2 \pi m}\] where \(q\) is the charge of the proton, \(B\) is the magnetic field, and \(m\) is the mass of the proton. The period, \(T\), is just the reciprocal of the cyclotron frequency: \[T = \frac{1}{f_c}\] Now we can find the period of the proton's motion using its cyclotron frequency and the given magnetic field.
03

Calculate the pitch of the helical path

The pitch, \(p\), can be found by multiplying the parallel velocity component by the period: \[p = v_{\parallel}T\] Now, we'll substitute the values we found in Steps 1 and 2 to calculate the pitch. Finally, we will convert the pitch from meters to centimeters by multiplying by 100.

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 which type of materials the magnetic susceptibility does not depend on temperature? (A) Diamagnetic (B) Paramagnetic (C) Ferromagnetic (D) Ferrite

Two identical conducting wires \(A O B\) and \(C O D\) are placed at right angles to each other. The wire \(A O B\) carries an electric current \(I_{1}\) and \(C O D\) carries a current \(I_{2} .\) The magnetic field on a point lying at a distance \(d\) from \(O\), in a direction perpendicular to the plane of the wires \(A O B\) and \(C O D\), will be given by [2007] (A) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)\) (B) \(\frac{\mu_{0}}{2 \mu}\left(\frac{I_{1}+I_{2}}{d}\right)^{\frac{1}{2}}\) (C) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)^{\frac{1}{2}}\) (D) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}+I_{2}\right)\)

Two particles \(X\) and \(Y\) having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describe circular paths of radii \(R_{1}\) and \(R_{2}\), respectively. The ratio of masses of \(X\) and \(Y\) is (A) \(\left(\frac{R_{1}}{R_{2}}\right)^{1 / 2}\) (B) \(\frac{R_{2}}{R_{1}}\) (C) \(\left(\frac{R_{1}}{R_{2}}\right)^{2}\) (D) \(\left(\frac{R_{1}}{R_{2}}\right)\)

Relative permittivity and permeability of a material \(\varepsilon_{r}\) and \(\mu_{r}\) respectively. Which of the following value of these quantities are allowed for a diamagnetic material? (A) \(\varepsilon_{r}=0.5, \mu_{r}=1.5\) (B) \(\varepsilon_{r}=1.5, \mu_{r}=0.5\) (C) \(\varepsilon_{r}=0.5, \mu_{r}=0.5\) (D) \(\varepsilon_{r}=1.5, \mu_{r}=1.5\)

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) \(n^{2} B\)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Torque and Rotational Motion

Read Explanation

Work Energy and Power

Read Explanation

Quantum Physics

Read Explanation

Astrophysics

Read Explanation

Kinematics 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.