At time t=0, an electron with kinetic energy 12 keV moves through x=0 in the positive direction

At time t=0, an electron with kinetic energy 12 keV moves...

At time $t=0$, an electron with kinetic energy $12 \mathrm{keV}$ moves through $x=0$ in the positive direction of an $x$ axis that is parallel to the horizontal component of Earth's magnetic field $\vec{B}$. The field's vertical component is downward and has magnitude $55.0 \mu \mathrm{T}$. (a) What is the magnitude of the electron's acceleration due to $\vec{B}$ ? (b) What is the electron's distance from the $x$ axis when the electron reaches coordinate $x=20 \mathrm{~cm} ?$

Key Concepts

Centripetal Acceleration

Centripetal acceleration is the acceleration directed toward the center of a circular path that an object experiences when moving in that circle. It is given by the formula a = v²/r, where v is the speed of the object and r is the radius of the circle. In the context of a charged particle in a magnetic field, the Lorentz force provides the necessary centripetal force to keep the particle in its circular path.

Kinetic Energy and Particle Velocity

The kinetic energy of a particle is related to its velocity through the equation K = 1/2 m v², where m is the mass of the particle. By knowing the kinetic energy, one can calculate the velocity, which is critical when analyzing the motion of charged particles in magnetic fields. This connection is essential for determining dynamic properties such as the radius of curvature and centripetal acceleration.

Lorentz Force

The Lorentz force is the fundamental force experienced by a charged particle moving in a magnetic field. It is mathematically expressed as F = q(v × B), where q is the charge, v is the velocity, and B is the magnetic field. This concept explains how magnetic fields can cause charged particles to deviate from a straight-line path, resulting in circular or helical trajectories depending on the angle between v and B.

Uniform Circular Motion

When a charged particle moves perpendicular to a uniform magnetic field, the constant magnitude of the Lorentz force acts as a centripetal force, causing the particle to undergo uniform circular motion. This means the particle maintains a constant speed while its direction continually changes, keeping it on a circular path. The radius of the circular motion depends on the particle’s momentum, charge, magnetic field strength, and mass.

Transcript

00:01 Here we're considering an electron who at time t equals zero moves through the point x equals zero with a kinetic energy of 12 kila electron volts, where the x -axis is parallel to the horizontal magnetic field of the earth, and we're told that the vertical component of the earth's magnetic field is 55 micro -teslas downward.

00:28 And we're asked to find the acceleration of the electron due to the earth's magnetic field, as well as its distance from the x axis when it reaches the location of x equals 20 centimeters.

00:47 So in order to find its acceleration, my mind always goes initially to newton's second law, and specifically that acceleration is equal to force over mass.

01:00 In this case, the force is being provided by the earth's magnetic field, and because it's moving horizontal, or sorry, parallel to the horizontal component, we only need to consider the vertical component because it will feel zero force from the horizontal component.

01:16 So our acceleration can be rewritten as qvb, over m because the force from a magnetic field that is perpendicular to your motion is going to be qvb.

01:35 In order to do this, we need to figure out what the velocity of the electron is to start, and we can do that by using the kinetic energy.

01:46 So it tells us initially that the kinetic energy is 12 kila electron volts.

01:51 I'm going to start by converting that to joules.

01:56 So to convert from kilo electron volts to joules, you're just going to multiply by the charge of an electron.

02:03 So we would have 12 kilojoules times e, the charge of an electron here.

02:12 But we also know that the kinetic energy is equal to 1 .5 mv squared.

02:19 So when you solve for the velocity here, you would get that v is equal to the square root of 2k over m, or in numbers, 6 .4968 times 10 to the 7 meters per second.

02:41 And sorry, i ran out of space there, so it's a little bit cramped.

02:45 But now we know the velocity so we can go back to our expression for acceleration, where q is going to be the charge of an electron, which is e...

Please give Ace some feedback