A proton is fired with a speed of 1.0×10 6 m/s through the parallel-plate capacitor shown in FIGURE P31.29. The capacitor's electric field is E =(1.0×10 5 V/m, down). How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?

Welcome back everyone. In this problem, an alpha particle passes deflected through a region with both electric and magnetic fields. This means that the net force on the particle is zero. An observer in the frame of the alpha particle would see the alpha particle stationary. How does the observer explain why the force on the alpha particle is zero? For our answer choices, it says the electric field is perpendicular to the magnetic field. Hence the forces cancel each other. B says there are no electric and magnetic forces on the alpha particle frame. C says the electric and the magnetic fields are in the same direction. Hence, they reinforce each other and cancel the forces. And D says the alpha particle has no charge. Hence there is no interaction with the electric and magnetic fields. No. In order to understand or in order to answer this question, we have to put ourselves in the place of the observer. So in other words, we are in the frame of the alpha particle and we want to understand why the force on the alpha particle seems to be zero or is zero. Rather. Now in this problem, we will assume that the region's electric and magnetic fields are uniform within the region where the alpha particle passes. And based on our responses here, we are trying to explain the force in terms of what's happening with the electric and magnetic forces and fields. So first, let's start to think about the electric field. OK. And for that, we'll need to think about the electric force. Now, for our electric force using the alpha particles reference frame, the electric field of the particle alpha would be equal to zero. And if the electric field is zero, that means the electric force would also be equal to zero. Now, if we put ourselves in the reference frame, this make makes sense because let's imagine that it's as if we were driving in a car, OK? And in a car while you're sitting in the car, everything else around you seems calm. It doesn't seem like any extra force is being exerted on you or anything different is happening while you're driving. But outside, you could obviously tell that our force is, you know, being exerted outside of the frame. Well, in the same way here because our alpha particle pass is un deflected. In other words, it's at rest then that tells us that our electric field, everything seems the same. Everything would. If you're in the reference frame, it doesn't seem like anything is changing, which is why our electric field is zero. And thus our electric force is zero. Now how about our magnetic force? What do we know? Well, given that our electric field is zero, our magnetic force can be given by the equation that says it's equal to the cross product of Q times V that is the charge and the velocity and b the strength of the magnetic field. Now, what do we know here? Well, recall, as we said, the alpha particle passes undeed it's assumed to be stationary. So that means is velocity then is assumed to be zero. And again, if we think about the car, that makes sense, let's say that you were in the car, right? And inside the car and you were looking at uh let's say you had a box somewhere on the dashboard of the car. OK. No, while you're driving in the car, the box is really moving at the same speed of the car. But to you, the observer inside the car, it doesn't look like the box is moving at all. So if we think about the car as the reference frame, then because you're in the reference frame, the the box looks stationary, that is the alpha particle would seem stationary. So that means if the velocity is as if the alpha particle is assumed to be stationary, the velocity is zero, which tells us then that the force is also going to be zero. So are electric and magnetic forces equal zero. Which means from the observer's point of view, there are no electric and magnetic forces on the alpha particles frame. If we look back on our answer choices, that would have been answer choice B thanks a lot for watching everyone. I hope this video helped.

Table of contents

Skip topic navigation

0. Math Review31m
- Math Review
  31m
1. Intro to Physics Units1h 29m
2. 1D Motion / Kinematics3h 56m
3. Vectors2h 43m
4. 2D Kinematics1h 42m
5. Projectile Motion3h 6m
6. Intro to Forces (Dynamics)3h 22m
7. Friction, Inclines, Systems2h 44m
8. Centripetal Forces & Gravitation7h 26m
9. Work & Energy1h 59m
10. Conservation of Energy2h 54m
11. Momentum & Impulse3h 40m
12. Rotational Kinematics2h 59m
13. Rotational Inertia & Energy7h 4m
14. Torque & Rotational Dynamics2h 5m
15. Rotational Equilibrium3h 39m
16. Angular Momentum3h 6m
17. Periodic Motion2h 9m
18. Waves & Sound3h 40m
19. Fluid Mechanics4h 27m
20. Heat and Temperature3h 7m
21. Kinetic Theory of Ideal Gases1h 50m
22. The First Law of Thermodynamics1h 26m
23. The Second Law of Thermodynamics3h 11m
24. Electric Force & Field; Gauss' Law3h 42m
25. Electric Potential1h 51m
26. Capacitors & Dielectrics2h 2m
27. Resistors & DC Circuits3h 8m
28. Magnetic Fields and Forces2h 23m
29. Sources of Magnetic Field2h 30m
30. Induction and Inductance3h 38m
31. Alternating Current2h 37m
32. Electromagnetic Waves2h 14m
33. Geometric Optics2h 57m
34. Wave Optics1h 15m
35. Special Relativity2h 10m

28. Magnetic Fields and Forces

Force on Moving Charges & Right Hand Rule

Physics

28. Magnetic Fields and Forces

Force on Moving Charges & Right Hand Rule

Previous problemNext problem

Problem 29c

Textbook Question

A proton is fired with a speed of 1.0×10⁶ m/s through the parallel-plate capacitor shown in FIGURE P31.29. The capacitor's electric field is E =(1.0×10⁵ V/m, down). How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?

Verified step by step guidance

Step 1: Begin by analyzing the problem in the proton's frame of reference. In this frame, the proton is stationary, and the parallel-plate capacitor appears to move past the proton at a velocity of 1.0×10^6 m/s.

Step 2: Recognize that the moving charged plates generate a magnetic field in addition to the electric field. This magnetic field is a result of the motion of the charges on the plates relative to the proton.

Step 3: Use the principle of relativity to understand that the proton experiences no net force. The electric field (E) exerts a downward force on the proton, but the magnetic field (B) generated by the moving plates exerts an upward force that exactly cancels the electric force.

Step 4: Apply the Lorentz force equation, F = q(E + v × B), to verify the cancellation of forces. In the proton's frame, the velocity (v) of the plates and the magnetic field (B) are such that the magnetic force balances the electric force.

Step 5: Conclude that the experimenter in the proton's frame explains the lack of force on the proton as a result of the perfect balance between the electric and magnetic forces acting on it.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above

Play a video:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electric Field

An electric field is a region around a charged particle where other charged particles experience a force. It is represented by electric field lines that indicate the direction and strength of the field. In this scenario, the electric field (E) is directed downward, indicating that a positive charge, like a proton, would typically experience a force in that direction.

Frame of Reference

A frame of reference is a perspective from which measurements and observations are made. In the proton's frame, it is moving with the same velocity as the proton, which means it perceives itself as stationary. This perspective is crucial for understanding why the proton does not experience a force from the electric field as the charged plates pass by.

Lorentz Force

The Lorentz force describes the force experienced by a charged particle moving through an electric and magnetic field. The force is given by the equation F = q(E + v × B), where q is the charge, E is the electric field, v is the velocity, and B is the magnetic field. In this case, since the proton is moving parallel to the electric field and in its own frame of reference, it does not experience a net force.

Watch next

Master Force on Moving Charges & Right Hand Rule with a bite sized video explanation from Patrick

Start learning

Related Videos

Related Practice

Textbook Question

A proton moves in the magnetic field B = 0.50 î T with a speed of 1.0 x 10⁷ m/s in the directions shown in FIGURE EX29.27. For each, what is magnetic force F on the proton? Give your answers in component form.

views

Textbook Question

An electron travels with $\vec{v} = (5.0 \times 10^{6} \hat{i}) m/s$ through a point in space where $\vec{E} = (2.0 \times 10^{5} \hat{i} - 2.0 \times 10^{5} \hat{j}) V/m$ and $\vec{B} = - 0.10 \hat{k} T$ . What is the force on the electron?

135

views

Textbook Question

What is the force (magnitude and direction) on the proton in FIGURE P31.28? Give the direction as an angle cw or ccw from vertical.

views

Textbook Question

A rocket cruises past a laboratory at 1.00×10⁶ m/s in the positive x-direction just as a proton is launched with velocity (in the laboratory frame) $\vec{v} = (1.41 \times 10^{6} \hat{i} + 1.41 \times 10^{6} \hat{j})$ m/s. What are the proton's speed and its angle from the y-axis in the rocket frame?

433

views

Textbook Question

Muons ( mass = 207 times the electron mass, same charge -e ) are accelerated horizontally by 65 kV. They then pass through a uniform magnetic field B for a distance of 3.8 cm, which deflects them upward so they reach a detector 22 cm away, 11 cm above their original direction. Estimate the value of B.

398

views

Textbook Question

A 3.40-g bullet moves with a speed of 155 m/s perpendicular to the Earth’s magnetic field of 5.00 x 10⁻^-5T. If the bullet possesses a net charge of 18.5 x 10^-9 C, by what distance will it be deflected from its path due to the Earth’s magnetic field after it has traveled 1.25 km?

453

views

Textbook Question

An electron moves with velocity $\vec{v} = (6.0 \hat{i} - 7.0 \hat{j}) \times 10^{4} m/s$ in a magnetic field $\vec{B} = (- 0.80 \hat{i} + 0.60 \hat{j}) T$ . Determine the magnitude and direction of the force on the electron.

221

views

Textbook Question

(II) An electron experiences the greatest force as it travels 2.8 x 10⁶ m/s in a magnetic field when it is moving northward. The force is vertically upward and of magnitude 7.4 x 10^-13 N. What is the magnitude and direction of the magnetic field?

409

views

Show more practices

Download the Mobile app

Privacy

Do not sell my personal information

Accessibility

Patent notice

Sitemap

A proton is fired with a speed of 1.0×106 m/s through the parallel-plate capacitor shown in FIGURE P31.29. The capacitor's electric field is E =(1.0×105 V/m, down). How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?

Key Concepts

Electric Field

Frame of Reference

Lorentz Force

Watch next

A proton is fired with a speed of 1.0×10⁶ m/s through the parallel-plate capacitor shown in FIGURE P31.29. The capacitor's electric field is E =(1.0×10⁵ V/m, down). How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?