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

T=0, an electron with kinetic energy 12 moves through...

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

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

Key Concepts

Lorentz Force

The Lorentz force is the fundamental force acting on a charged particle moving in a magnetic field, described by the equation F = q(v × B). This force is always perpendicular to both the velocity of the particle and the magnetic field, which causes the particle to follow a curved, typically circular, path without changing its speed.

Circular Motion in a Magnetic Field

When a charged particle enters a uniform magnetic field, the perpendicular Lorentz force results in circular motion. This concept is critical because the magnetic force acts as a centripetal force, determining parameters such as the radius of curvature and the trajectory of the particle.

Kinetic Energy and Velocity Relationship

The kinetic energy of a particle is related to its speed through the equation KE = ½mv². This relationship allows one to calculate the velocity of the particle, which is necessary for determining the magnitude of the Lorentz force and the radius of the circular motion in a magnetic field.

Centripetal Acceleration

Centripetal acceleration is the acceleration that keeps a particle moving along a circular path and is given by the formula a = v²/r. In the context of a charged particle moving in a magnetic field, the Lorentz force provides the necessary centripetal force that results in this acceleration.

Transcript

00:01 Hello student this is a problem from magnetic field and in this problem in part a we have to find the magnitude of electrons acceleration due to magnetic field we know that electric force is equal to q into v is the velocity into magnetic field we know that force is equal to m into a so the acceleration will be equal to q v b by m now we have to find the velocity because we know the charge magnetic field and mass but we don't know the velocity so we know that we have given kinetic energy we know that kinetic energy is equal to half mv square so from here we will get the velocity is under root 2k by m we will put the value of kinetic energy and mass kinetic energy is given to us which is 12 kilo electron volt i convert it into jule so i multiply with 1 .60 into 10 to the power minus 19 this is in jule divided by mass which is 9 .11 into 10 .10 to the power minus 31 kg so the velocity will be in s i unit which is meter per second 6 .49 into 10 to the power 7 meter per second this is the velocity now we will put this velocity here in this equation so we will get the acceleration which is equal to q which is 1 .6 into 10 to the power minus 19 velocity is 6 .49 into 10 to the power 7 meter per second magnetic field is given 55 into 10 to the power minus 6 tesla divided by again mass is 9 .11 into 10 to the power minus 31 kg so after solving this we will get the acceleration 6 .27 into 10 to power 14 meter per second square.

02:36 This is the value of acceleration.

02:40 Now in part b, we have to find in part v what is the electrons distance from the x -axis when the electrons reach coordinate x is equal to 20 centimeter.

02:52 Suppose we ignore any vertical deflection here.

02:58 So suppose it deplace is displaced by 20 centimeter...

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