∙An electron at point A in Figure 20.61 has a speed v0 of 1.41 ×10^6 m / s . Find (a) the magnit

∙An electron at point A in Figure 20.61 has a speed v0 of...

\bullet An electron at point $A$ in Figure 20.61 has a speed $v_{0}$ of $1.41 \times 10^{6} \mathrm{m} / \mathrm{s}$ . Find (a) the magnitude and direction of the magnetic field that will cause the electron to follow the semicircular path from $A$ to $B$ and (b) the time required for the electron to move from $A$ to $B$ . (c) What magnetic field would be needed if the particle were a proton instead of an electron?

Key Concepts

Lorentz Force

The Lorentz force is the force exerted on a charged particle moving in a magnetic field. It is given by the cross product of the velocity of the particle and the magnetic field, meaning the force is perpendicular to both the velocity and the field. This is the fundamental force that causes the charged particle to deviate from a straight-line path in a magnetic field.

Uniform Circular Motion of Charged Particles

When a charged particle moves in a uniform magnetic field, the Lorentz force acts as the centripetal force, resulting in uniform circular motion. The balance between the magnetic force and the required centripetal force allows for the derivation of the radius of the path of the particle, connecting its speed, charge, mass, and the magnetic field strength.

Cyclotron Frequency and Period

The cyclotron frequency is the frequency at which a charged particle orbits in a uniform magnetic field; it depends solely on the charge-to-mass ratio of the particle and the magnetic field strength. The period is the reciprocal of this frequency, representing the time it takes for one full revolution, which is crucial in analyzing the timing of the charged particle’s motion through a magnetic field.

Effect of Charge-to-Mass Ratio on Particle Motion

Different particles, such as electrons and protons, have different charge-to-mass ratios. This ratio affects the curvature of their path in a magnetic field since a lower charge-to-mass ratio leads to a larger radius for the same magnetic field strength and speed. Understanding this concept is essential when comparing the motion of different charged particles in identical magnetic conditions.

Transcript

00:01 So if we were to draw the trajectory of the electron, it's going to go through a half circle.

00:08 And so the velocity is always tangent to the curve.

00:15 And perpendicular to the velocity is always the magnetic force.

00:21 Here it would be going down.

00:26 And the velocity would be going straight to the right.

00:30 So here, the magnetic field must be directed into the page.

00:36 So we can label this.

00:42 So again, the magnetic field is always into the page.

00:45 And we know that the force is equaling mass times acceleration.

00:50 And the only force here would be the magnetic force.

00:55 And in this case, the acceleration is centripetal.

00:58 So we can say that the magnetic force equals the mass times the centripetal acceleration.

01:04 Or we can say that the absolute value of the charge times the velocity times the magnitude of the magnetic field times sine of phi is going to be equal to mv squared divided by r now here we know that phi is 90 degrees so this term is going to be equal to one it can be eliminated and we know that the magnitude of the magnetic field is going to be equal to the mass times the velocity divided by the absolute value of of the charge times the radius.

01:36 So at this point for part a, when we're trying to find b, we can simply say that this is going to be equal to 9 .11 times 10 to the negative 31st kilogram, and then the velocity of the electron is 1 .41 times 10 to the 6 meters per second.

01:56 And this will be divided by the charge of an electron, 1 .6 times 10 to the negative 19.

02:05 Kulams and then times the radius of 0 .05 meters.

02:15 And we find that the magnitude of the magnetic field is equaling 1 .61 times 10 to the negative 4th tesla's.

02:23 So this will be your answer for part a.

02:29 So that'll be your answer for a day.

02:31 And then for part b, they're asking us for a time...

Please give Ace some feedback