Question

In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-called cathode rays. One remarkable series of experiments with cathode rays, conducted by J. J. Thomson around 1897, led to the discovery of the electron.

With the idea that cathode rays were charged particles, Thomson used a cathode-ray tube to measure the ratio of charge to mass, q/m, of these particles, repeating the measurements with different cathode materials and different residual gases in the tube.

Part A

What is the most significant conclusion that Thomson was able to draw from his measurements?

He found a different value of q/m for different cathode materials.

He found the same value of q/m for different cathode materials.

From measurements of q/m he was able to calculate the charge of an electron.

From measurements of q/m he was able to calculate the mass of an electron.

Part B

What is the distance Δy between the two points that you observe? Assume that the plates have length d, and use e and m for the charge and the mass of the electrons, respectively.

Express your answer in terms of e, m, d, v0, L, and E0.

Part C

Now imagine that you place your entire apparatus inside a region of magnetic field of magnitude B0 (Figure 2) . The magnetic field is perpendicular to E⃗ 0 and directed straight into the plane of the figure. You adjust the value of B0 so that no deflection is observed on the screen.

What is the speed v0 of the electrons in this case?

Express your answer in terms of E0 and B0.

Part D

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the plates (with no magnetic field present). When you add the magnetic field as described in Part C, to what value do you have to adjust its magnitude B0 to observe no deflection?

Assume that the plates are 6.00 cm long and that the distance between them and the screen is 12.0 cm.

Express your answer numerically in tesla.

Part E

Now suppose you carry out a second Thomson experiment with a different beam that contains two types of particles. In particular, both types have the same mass m as an electron, but one has charge e and the other has charge 2e. The beam is filtered, such that both types of particle have the same speed. As in the previous experiment, initially only the electric field is imposed and the deflection of the beam is observed as a spot on the screen; then, in the second phase of the experiment, one attempts to tune the magnetic field to exactly cancel the effect of the electric field.

What would you observe on the screen during this experiment?

Two off-centered spots in both the first and the second phases of the experiment

Two off-centered spots in the first phase of the experiment; one centered spot in the second phase of the experiment

Two off-centered spots in the first phase of the experiment; one centered and one off-centered spot in the second phase of the experiment

One off-centered spot in the first phase of the experiment; one centered and one off-centered spot in the second phase of the experiment

One off-centered spot in the first phase of the experiment; two off-centered spots in the second phase of the experiment

One off-centered spot in both the first and the second phases of the experiment

Answer 1

Concepts and reason

The concept required to solve the given question are the equation of kinematics, relation between the electric field and magnetic field, magnetic force and electric force.

First, find out the most significant conclusion of the Thomson’s experiment from the given set of choices. In the next part, use the relation between the distance, velocity and time to find the time taken by the charge to travel through the electric field. Then, use equation of kinematics and the relation between the acceleration and electric field to find the vertical distance between the two points.

In the third section, use the expressions for electric and magnetic forces and equate them to find the speed of the electrons. In the next part, rearrange the expression obtained for the vertical distance between the points them to arrive at an expression for the speed. Substitute the values given in the question and find the speed. Applying the relation between the electric field, magnetic field and speed; the magnetic field that produces zero deflection can be calculated.

In the last section, use the conclusions from the experiment and arrive at the conclusion for the second phase of experiment that to cancel the effect of electric field by tuning the magnetic field.

Fundamentals

Distance can be explained as a scalar quantity, which refers to “how much ground an object has covered during its motion”. Distance is a scalar quantity.

Speed of the object is the “rate of change of distance moved by the object”. It is a scalar quantity. But velocity is a vector quantity. It has both magnitude and direction. Velocity of the object is the rate of change of displacement.

Acceleration of an object is defined as the “rate of change of velocity”. It is also a vector quantity.

Equations of kinematics are the set of relations that can be applicable to constant velocity motion or constant acceleration motion. If the acceleration is changing in a motion, then these relations cannot be applicable.

The kinematic equation connecting the acceleration, time, initial and final velocity is,

Here, is the initial velocity, is the acceleration, is the time taken and is the final velocity.

The displacement of the particle is given by,

Here, is the displacement of the particle.

Magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the velocity of the charge and the magnetic field. It can be calculated as the “charge times the vector product of the velocity and the magnetic field.

Here, is the magnetic force acting on the electron, q is the charge, v is the velocity of the electron and B is the magnetic field.

Newton’s second law states that the acceleration of an object produced by a net force is directly proportional to the magnitude of net force and inversely proportional to the mass of the object.

Mathematical expression for Newton’s law is given by,

$F=ma $

Here, F is the force acting, m is the mass and a is the acceleration.

Electric force is the attractive or repulsive interaction between two charged objects.

Expression for electric force is,

Here, q is the charge, and E is the electric field produced.

(A)

In the Cathode ray tube experiment, J.J. Thomson used a heated cathode as the source of the charged particle. These charged particles were known as the cathode rays. He tried to prove that these charged particles are negatively charged particles which are called electrons.

Here, the same particles are produced even if cathode materials are different. Their mass and charge are the same. Thus the charge to mass ratio of these negatively charged particles are same.

Hence the conclusion is that he found the same value of for different cathode materials.

(B)

Consider the expression for the horizontal displacement of the charged particle,

$d=voto $

Here, is the displacement, is the initial velocity and is the time taken by the charge to travel through the electric field.

Time taken by the charge is,

…… (1)

Applying the kinematics equation of motion,

Velocity of the electron along the vertical direction,

From Newton’s second law of motion,

$F=ma $ = …… (2)

Electric force acting on the particle is,

…… (3)

Here, is the electric field and is the charge.

Comparing equations (2) and (3),

$F=F. $

Both the forces are equal. This can be written as,

For an electron; the charge is $q=e $.

Thus the acceleration of the particle is,

…… (4)

Initial velocity of the particle in the vertical direction is zero. Substituting equations (1), (4) in the expression for the vertical velocity,

$+6=1 $ …… (5)

Kinematic equation for displacement is,

Consider the expression for vertical displacement of the particle during the time ,

Here, is the time taken for covering the distance by the electron.

Substitute for , for and for .

…… (6)

Horizontal displacement of the particle during the time ,

Hence the time taken by the charge to cover the distance is,

…… (7)

Use the expression for distance in terms of velocity and time,

$Ay-y=v,ti $

Distance between the two observed points is,

Substitute equation (5), (6) and (7) in the above expression,

(C)

Here, the deflection is zero. This means that the electric and magnetic forces acting on the particle exactly cancel each other.

That is,

Magnitude of the magnetic force is,

In this case, the magnetic field is perpendicular to the velocity vector.

Substitute for . for and for .

Electric force acting on the electron,

Equating both of the forces,

(D)

The distance between the two observed points is,

Rearrange the above equation,

Substitute for , for , for , for , for and for .

Using the relation,

Magnitude of the magnetic field is,

Substitute for and for .

(E)

Here, there are two particles are chosen such that both are having the mass of electron, and same speed. But one particle is having a charge of and the other is having a charge .

In the first phase of the experiment, both the particles experience an electric force along the vertical direction.

Acceleration of the particle is given by,

Here, the acceleration and the deflection of the particles are different. Hence there would be two off centered spots.

During the second phase of experiment, the magnetic field exactly cancels the electric field. Thus the electric and magnetic force cancel each other. The net force acting on the particle is zero. There is no deflection and the particles travel straight to the screen. Hence, there is only one centered spot.

Since, the forces get cancel each other, the particles cannot produce two off centered spots on the screen in the second phase of experiment. Also, since the charges are different, their deflection is different. Hence it cannot produce one off centered spot during the first phase of experiment. There will be two off centered spots on the screen.

Ans: Part A

The significant conclusion of Thomson’s experiment is that: He found the same value of for different cathode materials.

Part B

The distance between the observed points is .

Part C

The expression for the speed of the electron is .

Part D

The magnitude of magnetic field is .

Part E

The result of the experiment with different particles is two off centered spots in the first phase of experiment and one centered spot in the second phase of experiment.