Question

An electron moves at speed 5.6 x 106 m/s toward the velocity selector shown in (Figure 1)

 An electron moves at speed 5.6 x 106 m/s toward the velocity selector shown in (Figure 1). A 0.12-T magnetic field points into the paper.

image.png

 Part A

 Determine the magnitude of the magnetic force that the magnetic field exerts on the electron.

 Part B

 Determine the direction of the magnetic force that the magnetic field exerts on the electron.

  •  upward

  •  downward

 Part C

 What E field magnitude is required so that the electric force exerted on the electron is equal in magnitude and opposite in direction to the magnetic force?

 Part D

 What E field direction is required so that the electric force exerted on the electron is equal in magnitude and opposite in direction to the magnetic force?

  •  upward

  •  downward

3 0
Add a comment Improve this question Transcribed image text
Answer #1

Concept: use the magnetic force on a moving charged particle and then equate the electric force with the magnetic force to find out the magnitude of the electric field


***************************************************************************************************
Check the answer and let me know immediately if you find something wrong... I will rectify the mistakes asap if any

Add a comment
Know the answer?
Add Answer to:
An electron moves at speed 5.6 x 106 m/s toward the velocity selector shown in (Figure 1)
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
  • Velocity Selector-multiple choice

    The Velocity SelectorIn experiments where all the charged particles in a beam are required to have the same velocity (for example, when entering a mass spectrometer), scientists use a velocity selector. A velocity selector has a region of uniform electric and magnetic fields that are perpendicular to each other and perpendicular to the motion of the charged particles. Both the electric and magnetic fields exert a force on the charged particles. If a particle has precisely the right velocity, 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...

  • 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

  • A proton moves to the left with a speed of 1.25 · 106 m/s into a...

    A proton moves to the left with a speed of 1.25 · 106 m/s into a region of constant magnetic field as shown in the figure below. The magnitude of the magnetic field is 1.91 · 10-2 T and the direction of the field is into the page. When the proton initially enters the magnetic field, the magnitude of the force on the proton (in N) is A: 2.160×10-15 B: 2.872×10-15 C: 3.820×10-15 D: 5.081×10-15 E: 6.757×10-15 F: 8.987×10-15 G:...

  • 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...

  • An electron moves with speed 5.0 times 10^5m/s in a uniform magnetic field of strength 0.60...

    An electron moves with speed 5.0 times 10^5m/s in a uniform magnetic field of strength 0.60 T that points Hast. At some instant, the electron experiences an upward magnetic force of magnitude 4.0 times 10^14 N. In what direction is the electron moving at that instant? [If there is more than one possible direction, describe all the possibilities.] On a carefully drawn and labeled diagram, show the (possible) direction(s) of the electron's velocity relative to the magnetic field and the...

  • 1. An electron moves with a speed of 1.00 x 10^5 m/s through planet Spiro’s magnetic...

    1. An electron moves with a speed of 1.00 x 10^5 m/s through planet Spiro’s magnetic field, which has a value of 55.0 x 10^-6 T at a particular location. When the electron moves westward, the magnetic force acting on it is directed straight upward, and when it moves northward, no magnetic force acts on it. a. What is the direction of the magnetic field? Explain your reasoning. b. What is the strength of the magnetic force when the electron...

  • Question 1 (10 points). (a) Determine the direction of the force on the moving charges described ...

    practice exam help. will upvote Question 1 (10 points). (a) Determine the direction of the force on the moving charges described below Magnetic field into paper, positive charge moving to the right Magnetic field and velocity vectors as shown with the charge being negative xxx부x (b) An electron moves in the platnve of this paper toward the top of the page. A magnetic field is also in the plane of the page and directed toward the right. The direction of...

  • An electron at point A in the figure shown has a speed of vo = 1.41 x 106 m/s.

    An electron at point A in the figure shown has a speed of vo = 1.41 x 106 m/s. a) Find the magnitude (15 pts) and direction (5 pts) of the magnetic field that will cause the electron to follow the path shown from A to B. b) What is the time required for the electron to move from A to B?

  • An electron and a proton are each moving at 845 km/s in perpendicular paths as shown in the figure(Figure 1)

    An electron and a proton are each moving at 845 km/s in perpendicular paths as shown in the figure(Figure 1) . At the instant they are at the positions shown in the figure. (A) Find the magnitude of the total magnetic field they produce at the origin. (B) Find the direction of the total magnetic field they produce at the origin. (C) Find the magnitude of the magnetic field the electron produces at the location of the proton. (D)Find the direction of the magnetic field...

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