Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 37: Q96P (page 1152)

Question:A 2.50 MeV electron moves perpendicularly to a magnetic field in a path with a 3.0 cm radius of curvature. What is the magnetic field B? (See Problem 53)

Short Answer

Expert verified

The magnetic field for the electron is 0.178T.

Step by step solution

01

Identification of given data

The energy of electron is E=2.50MeV

The radius of curvature is r=3cm

The magnetic field for the electron is found by equating the necessary centripetal force by magnetic force on the electron.

02

Determination of magnetic field for electron

The magnetic field for electron is given as:

B=2mEqr

Here, q is the charge of electron and its value is 1.6×10-19C , m is the mass of electron and its value is 9.1×10-31kg

Substitute all the values in the above equation.

B=29.1×10-31kg2.50MeV1.6×10-13J1MeV1.6×10-19C3cm10-2m1cmB=0.178t

Therefore, the magnetic field for the electron is 0.178T.

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

In the reaction p+F19→α+O16, the masses are m(p) = 1.007825 u, m(a) = 4.002603 u, m(F) = 18.998405 u, m(O) = 15.994915 u. Calculate the Q of the reaction from these data.

The total energy of a proton passing through a laboratory apparatus is 10.611-nJ. What is its speed parameter β ? Use the proton mass given in Appendix B under “Best Value,” not the commonly remembered rounded number.

Figure 37-16 shows a ship (attached to reference frame S') passing us (standing in reference frame S) with velocity v→=0.950ci^. A proton is fired at speed 0.980c relative to the ship from the front of the ship to the rear. The proper length of the ship is 760 m . What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to (a) a passenger in the ship and (b) us? Suppose that, instead, the proton is fired from the rear to the front. What then is the temporal separation between the time it is fired and the time it hits the front wall according to (c) the passenger and (d) us?

A space traveler takes off from Earth and moves at speed 0.9900ctoward the star Vega, which is 26.00lydistant. How much time will have elapsed by Earth clocks (a) when the traveler reaches Vega and (b) when Earth observers receive word from the traveler that she has arrived? (c) How much older will Earth observers calculate the traveler to be (measured from her frame) when she reaches Vega than she was when she started the trip?

A spaceship whose rest length is 350 m has a speed of 0.82c with respect to a certain reference frame. A micrometeorite, also with a speed of 0.82c in this frame, passes the spaceship on an antiparallel track. How long does it take this object to pass the ship as measured on the ship?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Kinematics Physics

Read Explanation

Quantum Physics

Read Explanation

Radiation

Read Explanation

Electricity

Read Explanation

Further Mechanics and Thermal 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.

Sign-up for free