Question

An electron moves at 2.50*106 m/s through a region inwhich there is a magnetic field of...

An electron moves at 2.50*106 m/s through a region inwhich there is a magnetic field of unspecified direction andmagnitude 7.40*10-2 T. (a) What are the largest andsmallest possible magnitudes of the acceleration of the electrondue to the magnetic field? (b) If the actual acceleration of theelectron is one-fourth of the largest magnitude in part (a), whatis the angle between the electron velocity and the magneticfield?
0 0
Add a comment Improve this question Transcribed image text
Answer #1
Concepts and reason

The concepts required to solve the given problem are magnetic field and magnetic force.

First calculate the largest and smallest possible magnitudes of the acceleration of the electrons due to the magnetic field and then the angle between the electron velocity and the magnetic field can be calculated.

Fundamentals

Magnetic field is defined as the region around the permanent magnet where the effects of magnetic force can be experienced. Magnetic fields are produced by moving charges or current flowing in a conductor.

The SI unit of magnetic field is Tesla.

The magnitude of magnetic field is calculated by using the magnetic parts of the Lorentz force, which is given as follows:

#magnetice = qlüxB)

Here, q is the charge, v is the velocity of the moving charge, and B is the magnitude of magnetic field.

(a)

The magnetic force exists on the electron is,

#magnetice = qlüxB)

Here, q is the charge, v is the velocity of the moving charge, and B is the magnitude of magnetic field.

Rewrite the above expression as follows:

magnetic = qvBsin 0

Here, is the angle between velocity vector and magnetic field.

Substitute 90°
for and mamax
for magnetic
in the expression as follows:

F =9(VB)

UU
(916) = strany

Here, amax is the maximum acceleration and m is the mass of electron. Maximum acceleration occurs when the force acting on the charge particle is maximum. When value of sin θ
is equal to 1.

Substitute (3 6-019*1)
for q, (2.50x109 m/s)
for v, (7.40x10? T)
for B, and (8418-01*16)
for m in expression as follows:

qvB
amax
m
((1.6x10-1° C)(2.50x10^ m/s)(7.40x10^2 T)).
9.11x10-” kg
= 3.25 x 106 m/s?

(b)

The magnitude of new acceleration is,

al-qvB sin e

Here, q is the charge, v is the velocity of the moving charge, m is the mass of electron, a’ is the actual acceleration, is the angle between velocity vector and magnetic field, and B is the magnitude of magnetic field

Substitute (3.25x10 m/s
4
for and 3.25x10
m/s
for qvB
in above expression:

a=qvB sin 
(3.25*10* m52) = (325–10“ mxs*)sino
sino =
O= 14.47°

Ans: Part a

The magnitude of maximum and minimum acceleration is (3.25x10 m/s)
and 0 m/s2.

Part b

The angle between the electron velocity and the magnetic field is 14.479
.

Add a comment
Know the answer?
Add Answer to:
An electron moves at 2.50*106 m/s through a region inwhich there is a magnetic field of...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • An electron moves at 3.00×106 m/s through a region in which there is a magnetic field...

    An electron moves at 3.00×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.90×10−2 T . a- What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? b- What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field? c- If the actual acceleration of the electron is 14 of the largest magnitude in part (a), what is the...

  • PART A: An electron moves at 3.00×106 m/s through a region in which there is a...

    PART A: An electron moves at 3.00×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.50×10−2 T . A1: An electron moves at 3.00×106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.50×10−2 T . A2: What is the smallest possible magnitude of the force on the electron due to the magnetic field? Express your answer in newtons to the nearest integer. A3: If...

  • Exercise 27.6 What is the largest possible magnitude of the acceleration of the electron due to...

    Exercise 27.6 What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? An electron moves at 2.80*106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.60×10-2 T m/s? Submit My Answers Give Up Part B What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field? a" m/s Submit My Answers Give Up Part C "the actu" acceleration of...

  • An electron moves through a region with a uniform magnetic field of unknown magnitude directed in...

    An electron moves through a region with a uniform magnetic field of unknown magnitude directed in the +y direction At the instant that the electron is moving in the +z direction with a velocity of 2.86e6 m/s, it experiences a magnetic force of 1.24 pN in the +x direction. Calculate the magnitude of the magnetic field

  • A proton moving at 7.60 106 m/s through a magnetic field of magnitude 1.76 T experiences...

    A proton moving at 7.60 106 m/s through a magnetic field of magnitude 1.76 T experiences a magnetic force of magnitude 7.40 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest. Enter only positive values between 0 and 360.) smaller value ° larger value °

  • An electron in a TV camera tube is moving at 5.20×106 m/s in a magnetic field...

    An electron in a TV camera tube is moving at 5.20×106 m/s in a magnetic field of strength 57 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? a) Maximum force? b) Minimum force? c) At one point the acceleration of the electron is 4.316×1016 m/s2. What is the angle between the electron velocity and the magnetic field? (deg)

  • An electron in a TV camera tube is moving at 7.20×106 m/s in a magnetic field...

    An electron in a TV camera tube is moving at 7.20×106 m/s in a magnetic field of strength 65 mT. Without knowing the direction of the field, what can you say about the greatest and least magnitude of the force acting on the electron due to the field? Maximum force? Submit Answer Tries 0/99 Minimum force? Submit Answer Tries 0/99 At one point the acceleration of the electron is 7.509×1016 m/s2. What is the angle between the electron velocity and...

  • A proton moving at 4.50 106 m/s through a magnetic field of magnitude 1.80 T experiences...

    A proton moving at 4.50 106 m/s through a magnetic field of magnitude 1.80 T experiences a magnetic force of magnitude 7.80 10-13 N. What is the angle between the proton's velocity and the field? (Enter both possible answers from smallest to largest.) a. in ° b. in °

  • An electron that has velocity (3.4 x 10 m/s)i+ (2.5 x 106 m/s)j moves through a...

    An electron that has velocity (3.4 x 10 m/s)i+ (2.5 x 106 m/s)j moves through a magnetic field B (0.03 Ti (0.15 T)]. (a) Find the force on the electron magnitude X* N direction Etera mumber (b) Repeat your calculation for a proton having the same velocity magnitude direction

  • An electron moves in a region where the magnetic field is constant and has a magnitude...

    An electron moves in a region where the magnetic field is constant and has a magnitude of 3.0 x 10-4 T. If the magnitude of the acceleration of the electron is 8.0x1012m/s2 at an instant when the speed is 200 km/s, what is the angle between the velocity and the field. Restrict your answer to the first quadrant values (from 0 to 90o). Do not write the symbol for degrees. Round off your answer to one decimal place. Show Work.

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT