A proton is fired with a speed of 1.0 ×10^6 m / s through...

A proton is fired with a speed of $1.0 \times 10^{6} \mathrm{m} / \mathrm{s}$ through the parallel-plate capacitor shown in Figure 35.31. The capacitor's electric field is $\vec{E}=\left(1.0 \times 10^{5} \mathrm{V} / \mathrm{m}, \text { down }\right)$ a. What magnetic field $\overrightarrow{\boldsymbol{B}}$, both strength and direction, must be applied to allow the proton to pass through the capacitor with no change in speed or direction? b. Find the electric and magnetic fields in the proton's reference frame. c. How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by? (FIGURE CAN'T COPY)

A proton is fired with a speed of $1.0 \times 10^{6} \mathrm{m} / \mathrm{s}$ through the parallel-plate capacitor shown in Figure 35.31. The capacitor's electric field is $\vec{E}=\left(1.0 \times 10^{5} \mathrm{V} / \mathrm{m}, \text { down }\right)$
a. What magnetic field $\overrightarrow{\boldsymbol{B}}$, both strength and direction, must be applied to allow the proton to pass through the capacitor with
no change in speed or direction?
b. Find the electric and magnetic fields in the proton's reference frame.
c. How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?
(FIGURE CAN'T COPY)

Key Concepts

Reference Frame Analysis

Reference frame analysis is the study of physical phenomena as observed from different inertial frames. In the context of electromagnetism, shifting to the particle’s rest frame alters the observed electric and magnetic fields, yet the net force on the particle remains zero if the fields were originally balanced in the lab frame. This concept underscores the consistency of physical laws in all inertial frames and helps in reconciling different observations with the principle of relativity.

Electromagnetic Field Transformation

When analyzing electromagnetic interactions from different inertial frames, it is necessary to transform the electric and magnetic fields according to the rules of special relativity. Field transformations show that what appears purely as an electric field in one frame might be seen as a combination of electric and magnetic fields in another. Understanding these transformations is crucial in explaining observations, such as a particle experiencing no net force in its rest frame even as fields appear to change.

Magnetic Force Cancellation

Magnetic force cancellation involves choosing a magnetic field such that the magnetic force (v × B) exactly counteracts the force due to an existing electric field. For a particle moving at a known velocity, the direction and magnitude of the magnetic field can be precisely determined to cancel the electric force, resulting in no net force. This approach is important in controlling particle trajectories in devices like particle accelerators and mass spectrometers.

Lorentz Force

The Lorentz force law is central to understanding the motion of charged particles in electromagnetic fields. It states that a charged particle experiences a force due to electric and magnetic fields given by F = q(E + v × B). This concept is vital as it explains how the simultaneous presence of electric and magnetic fields can be arranged so that their forces cancel, allowing a particle to move undeflected.

Transcript

00:01 Question 31 is this shown with this figure here, where a proton is fired at the velocity given through a parallel plate capacitor, and we're told the electric field is pointing downward with this magnitude.

00:13 So we're going to find out a couple of things for part a.

00:17 We want to find what magnetic field for both strength and direction must be applied to allow the proton to pass through without changing its speed or direction.

00:25 So what magnetic field must exist such that this proton just goes straight.

00:31 Sorry, i'm going to a different color here.

00:35 Just go straight through and seemingly is unaffected by the electric field.

00:41 So because the force due to the electric field will be pushing it downward, that means we know that the force on the magnetic field must be acting upward.

00:52 So we know that b, in order for that to occur, because again, it's a cross product of, if you think the force due to b, and given my qv cross b.

01:03 In order for the force to be acting in the upward direction and we know the particles traveling to the right that means the magnetic field b must be into the page first of all so we can determine the direction automatically b is into the page great so in order for this to occur we know that the force due to the electric field must equal in magnitude the force in the magnetic field those two balance we have the force due to electric fields just charge times and for magnetic field, i already kind of spoiled it.

01:40 It's qv cross b.

01:44 Our charges cancel out.

01:45 So we can just solve for the magnetic field to be the electric field divided by the velocity of our charge.

01:53 And so we can go ahead and just plug those values in.

01:56 And you can find that the magnetic field turns out, just by looking at these values now, it's going to be 0 .1 tesla.

02:15 And again, that's into the page based on, the right -hand rule of needing the magnetic force to be pushing the charge upward.

02:24 Great.

02:25 Part b, we want to find the electric and magnetic field in the protons reference frames.

02:31 So in the protons reference frames, it's the one we think of, it's the prime, it's the one moving along with the charge.

02:39 So in the scenario, it's the electric field from the outside point of view plus v -cross -b.

02:53 And so if we have our electric field, from our point of view, sorry, from the protons point of view, the electric field is, of course, pointing in the negative i direction, or j downward, which we can call negative j, but i didn't label it that way.

03:12 So it's negative plus v cross b.

03:17 That's a positive term.

03:20 So v is 1 .0 times 10 to the 6.

03:24 We know they're perpendicular to each other, so i don't need to worry about the cross product multiplied by b .1...

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