In a CRT, electrons moving at 1.8 ×10^7 m / s pass between the poles of an electromagnet where t

In a CRT, electrons moving at 1.8 ×10^7 m / s pass between...

In a CRT, electrons moving at $1.8 \times 10^{7} \mathrm{m} / \mathrm{s}$ pass between the poles of an electromagnet where the magnetic field is $2.0 \mathrm{mT}$ directed upward. (a) What is the radius of their circular path while in the magnetic field? (b) The time the electrons spend in the magnetic field is $0.41 \mathrm{ns} .$ By what angle does the direction of the beam change while it passes through the magnetic field? (c) In what direction is the beam deflected, as viewed by an observer louking al the screen?

In a CRT, electrons moving at $1.8 \times 10^{7} \mathrm{m} / \mathrm{s}$ pass between the poles of an electromagnet where the magnetic field is $2.0 \mathrm{mT}$ directed upward. (a) What is the radius of their circular path while in the magnetic field?
(b) The time the electrons spend in the magnetic field is $0.41 \mathrm{ns} .$ By what angle does the direction of the beam change while it passes through the magnetic field? (c) In what direction is the beam deflected, as viewed by an observer louking al the screen?

Key Concepts

Right-Hand Rule

The right-hand rule is a mnemonic used to determine the direction of the force exerted on a positive charge moving in a magnetic field. For electrons, which are negatively charged, the force direction is opposite to that predicted by the rule. This concept helps in visualizing and predicting the direction in which charged particles will be deflected when subjected to a magnetic field.

Angular Deflection

Angular deflection refers to the change in the direction of the particle's velocity as it moves along a curved path in a magnetic field. It is determined by the rate of rotation (angular frequency) and the time the particle spends in the magnetic field. This concept is essential in applications where precise control of beam direction is required.

Lorentz Force

The Lorentz force is the force acting on a charged particle moving in an electromagnetic field. It is given by the vector cross product F = q(v x B), which means the force is perpendicular to both the velocity of the particle and the magnetic field. This principle is fundamental in predicting the motion of charged particles when subjected to magnetic fields.

Centripetal Force

Centripetal force is the inward force required to keep an object moving in a circular path. When a charged particle moves through a magnetic field, the Lorentz force acts as the centripetal force, causing the particle to undergo circular motion. This relationship allows one to calculate parameters like the radius of the circular path.

Cyclotron Motion

Cyclotron motion describes the circular movement of charged particles under the influence of a perpendicular magnetic field. This motion is characterized by a constant magnitude of velocity and a well-defined radius and frequency of revolution, independent of the speed, which is crucial when calculating the curvature of the particle's path.

Transcript

00:01 In the given problem, speed of the electron in the given magnetic field, that is v is equal to 1 .8 into 10 to the power 7 meter per second and the magnetic field in which this electron is moving, that is having a magnitude of 2 .0 millie tesla and that is directed upward in the plane of paper now in the first part of the problem we have to find radius of the circular path described by the electron in this magnetic field and the expression for that radius is m into v by by bq, where m is the mass of the electron v is its speed b, the magnetic field and q, the charge over the electron.

01:29 So plugging in, all known values for the mass of the electron, that is 9 .1, into 10 -p.

01:36 Minus 31 kilogram multiplied by speed, which is given as 1 .8, into 10 -dish -par 7 meter per second divided by magnetic field 2 .0 millitazla or 2 .0 into 10 dash per minus 3 tesla multiplied by charge over the electron which is 1 .6 into 10 dash per minus 19 coulon.

02:04 So finally it is calculated to be equal to 5 .1 into 10 dash per minus 2 meter or finally we can say this is 5 .1 centimeter which becomes the answer for the first part of this problem.

02:25 Now in the second part of the problem the time consumed by this electron within this magnetic field is given as 0 .41 nanosecond.

02:44 So, in this much time, how much angle will be deviated by its velocity vector we have to find it? so if we assume this to be the circular trajectory of the electron, in the beginning, suppose this is the position of this electron, and then in the given time, 0 .41 nanosecond, suppose the electron reaches here.

03:12 So we have to find the angle, this angle we have to find, theta.

03:20 So to find it first of all, we find the total time period of the electron in this circular motion to complete this circular motion.

03:36 And that is given as that is given by the expression 2xm by bq.

03:43 So again plugging in all known values here, this is two times of 3 .1 .1.

03:48 Multiplied by the mass of electron again this is 9 .1 into 10 dash per minus 31 kilogram divided by magnetic field 2 .0 into 10 dash per minus 3 tesla multiplied by the charge over electron 1 .6 into 10 dash per minus 19 koolang so finally this time taken by the electron in the magnetic field means in case there is a complete magnetic field.

04:20 Then the time period of the electron comes out to be 17 .9 into 10 dash per minus 9 second, which can also be written as 17 .9 nanosecond...

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