January 12, 2024

Electrons

Cross-Field

Cross Field


The region in which both electric and magnetic fields are applied simultaneously and
perpendicular to each other is called a cross-field. The cross-field is useful to determine the speed of the charged particle.

Determination of Specific Charge of Electron: J. J. Thomson’s Experiment

J. J. Thomson’s experiment is performed to determine the specific charge of the electron. The specific charge is defined as the charge per unit mass.

The experimental arrangement for J. J. Thomson’s experiment is shown in the figure below. It consists of an evacuated glass tube G cRwhich is provided with the source of electrons,
set up for electric and magnetic field and a screen S. The electric field is provided by the
set of plates A and B connected with a high tension source (H. T.) so that the upper plate
A positive and lower plate B negative. The magnetic field is applied within the shaded
circular region, perpendicular inward to the plane of the paper (perpendicular to the electric
field). In this way, a cross-field is set up in the path of the electron.

The electron produced from the source is accelerated by the potential applied in plate A. The accelerated electron enters into the region of cross-field with some velocity v. There are three possibilities.

(i) If both electric and magnetic field is not applied then the electron moves in a straight path and strikes the center of the screen O, producing a glow at that point.

(ii) If only the electric field is applied then the electron deviates in an upward direction and strikes at point P on the screen.

(iii) If only a magnetic field is applied then the electron deviates in a downward direction and strikes at point Q on the screen.

Figure: Experimental arrangement for determination of specific charge of an electron.


First of all, the magnetic field is applied. Then the electron moves in a circular path along CD and strikes point Q on the screen. In this condition, we can write,

Force on electron due to magnetic field = Centripetal force on a circular path
π‘œπ‘Ÿ, 𝐹𝐡 = πΉπ‘š
π‘œπ‘Ÿ, 𝐡𝑒𝑣 = π‘šπ‘£^2/r
π‘œπ‘Ÿ, 𝑒/m = v/Br ……….(1)

Equation (1) gives the value of the specific charge of the electron after knowing the value of
v, B, and r.


(1) Measurement of B

The magnetic field can be measured with the help of a flux meter.

(2) Calculation of speed of electron: v

To calculate the velocity of we have to apply the cross-field, i.e., electric and magnetic field
both simultaneously in such a way that the force due to the electric and magnetic field are equal.
In this situation, the electron again moves in straight and strikes at the center of screen O.

Here,
Force on electron due to magnetic field = Force on electron due to electric field
π‘œπ‘Ÿ, 𝐹B = 𝐹𝐸
π‘œπ‘Ÿ, 𝐡𝑒𝑣 = 𝑒𝐸
π‘œπ‘Ÿ, 𝑣 =𝐸/𝐡 … … … … … … (2)

By knowing the value of electric and magnetic fields we can calculate the velocity of
electrons. The electric field can be calculated by using the formula
,
𝐸 =𝑉/𝑑… … (3)

Where V is the p.d. between the two plates A and B and d is the plate separation.


(3) Measurement of the radius of the circular path: r

In the figure, O’ is the center of the circular path CD having radius r.
Here, CO’ = DO’ = r

Let, ∠CO’D = ΞΈ

In the figure, CO and DQ are tangent to the curve path CD. Let us produce DQ backward so
that it meets CO at point E.

From geometry, ∠OEQ = θ
In the figure, from O’CD, we can write,

πœƒ = 𝐢𝐷̂/π‘Ÿ … … (4)

Also, from ⊿EOQ,

tan πœƒ = 𝑂𝑄/𝐸𝑄
For small ΞΈ, tanΞΈ~ΞΈ
∴ πœƒ = 𝑂𝑄/𝐸𝑄 … … . (5)
Equating equations (4) and (5), we get,

𝐢𝐷̂/r=𝑂𝑄/𝐸𝑄

π‘Ÿ = 𝐢𝐹 Γ— 𝐸𝑄/𝑂𝑄 … . . (6)

The quantity, 𝐢𝐷̂ β‰… 𝐢𝐹, diameter of the ring of the magnetic field.

Hence, by measuring the value of CF, EQ, and OQ directly in the experiment, the radius of
the circular path of an electron can be measured.

By putting the value of B, v, and r in equation (1), we can measure the value of the specific
charge of the electron (e/m). J. J. Thomson measured the value of the specific charge of
electron 1.76 Γ— 1011 C/kg in his experiment.

Mass of the Electron: me
Millikan’s oil drop experiment gives the value of the charge of an electron and J. J.
Thomson’s experiment gives the specific charge of the electron. Dividing the charge of
an electron by a specific charge, we get,

𝑒/π‘’β„π‘š = 1.6 Γ— 10βˆ’19 𝐢/1.76 Γ— 1011 𝐢/π‘˜π‘”
β‡’ π‘š = 9.1 Γ— 10βˆ’31 π‘˜π‘”

In this way, we can calculate the mass of an electron with the help of charge and specific
charge of electrons.

Millikan Oil Drop Experiment: Determination of Charge of Electron

R. A. Millikan set up this experiment for the determination of fundamental quantity in the
charge, i.e., the charge of electrons. This discovery won the Nobel Prize in 1923.

Working Principle

The working principle of this experiment is Stoke’s law of viscosity. Furthermore, the
motion of small spherical oil drops is observed in the presence of an electric field as well as in
the absence of an electric field.

Construction:

The experimental setup for Millikan’s oil drop experiment is shown in the figure. The experimental setup consists of a double-wall chamber inside which the experiment is performed. Two parallel plates are mounted inside the chamber which is connected with a high-tension source of p.d. 10,000 V, with the upper plate being positive. There is a small hole (H) in the upper plate, which is used to drop the oil drop. An atomizer is inserted at the upper right part which is used to spray the clock oil between the plates.

There are three windows in the chamber. One window is used to pass the light so that the oil drop inside it can be seen easily. The second window is used to pass the X–ray, which is necessary to ionize the oil drop so that it gets charged. The third window is placed at the front face for traveling microscope (T.M.), which is used to observe the oil drop and measure the terminal velocity.

Figure: Experimental arrangement for Millikan’s oil drop experiment.

Procedure
The experiment is performed in two steps:
(i) In the absence of an electric field
(ii) In the presence of electric field.

The oil drop is allowed to move between the plates with constant terminal velocity. The
time taken to move a certain distance is noticed to calculate the terminal velocity. The
terminal velocity in both steps is calculated.

Theory:

Let, ρ = density of the oil drop, 𝜎(air) = density of air, η = coefficient of viscosity of air, and r = radius of oil drop

In the absence of electric field oil drop moves under the effect of gravity and upthrust. Let
v1 be the terminal speed of oil drop in a downward direction such that the force due to
viscosity acts vertically upward.

F1 = 6πηrv1……….(1)

The weight of the oil drop is,

W = mg
β‡’ π‘Š = π‘‰πœŒπ‘”; 𝑉 = π‘£π‘œπ‘™π‘’π‘šπ‘’
β‡’ π‘Š =4/3πœ‹π‘ŸπœŒπ‘” … … (2)

The upthrust of air on the oil drop acts in the upward direction. The magnitude of upthrust is
equal to the weight of the displaced air by the oil drop.

Upthrust = weight of displaced air by drop

π‘ˆ =4/3πœ‹π‘ŸπœŽπ‘” … … . (3)
Resolving all forces, Newton’s second law gives,
βˆ‘πΉ = π‘œ
β‡’ π‘Š βˆ’ π‘ˆ βˆ’ 𝐹1 = 0
β‡’4/3πœ‹π‘ŸπœŒπ‘” βˆ’4/3πœ‹π‘ŸπœŽπ‘” = 6πηr𝑣1 … … . (4)
β‡’4/3πœ‹π‘Ÿ(𝜌 βˆ’ 𝜎)𝑔 = 6πηr𝑣1
β‡’ π‘Ÿ = √9Ξ·v1/2(𝜌 βˆ’ 𝜎)𝑔 … … . (5)

It gives the radius of the oil drop.
Now the electric field is applied with the upper plate being positive. The electric field is so
strong that the oil drop starts to move in an upward direction with constant speed. Let v2 be
the terminal speed of oil at this time such that the force due to viscosity acts vertically
downward.

F2 = 6πηrv2……….(6)
The force on oil drop having charge q in an electric field is given by,

FE = qE …….(7)

Resolving all forces, Newton’s second law gives,

βˆ‘πΉ = π‘œ
β‡’ 𝐹𝐸 + π‘ˆ βˆ’ 𝐹2 βˆ’ π‘Š = 0
β‡’ 𝐹𝐸 = 𝐹2 + (π‘Š βˆ’ π‘ˆ)
β‡’ π‘žπΈ = 6πηr𝑣2 + 4/3πœ‹π‘Ÿ(𝜌 βˆ’ 𝜎)𝑔
β‡’ π‘žπΈ = 6πηr𝑣2 + 6πηr𝑣1
β‡’ π‘ž =6πηr(𝑣2 + 𝑣1)/𝐸
β‡’ π‘ž =6πη(𝑣2 + 𝑣1)/𝐸√9Ξ·v1/2(𝜌 βˆ’ 𝜎)𝑔 … … . (8)

It gives the value of the electric charge contained in the oil drop. This experiment is repeated
again and again by taking the oil drop of different sizes and the charge in the oil drop is
calculated. It is found that each time the charge is found to be the integral multiple of
1.6 Γ— 10-19
,
i.e., q = n Γ—1.6 Γ— 10-19C

This value 1.6 Γ— 10-19 is the basic unit of charge which is the magnitude of the charge of one
electron.
From Millikan’s experiment, it is concluded that;

(i) The electronic charge is the most fundamental change in nature.
(ii) The charge in every matter can be expressed in terms of an integer multiple of the charge of the electron, i.e., q = ne. This is called the quantization of charge

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