1. This experiment aims to determine the specific charge (e/m) of electrons using the helical coil method. A cathode ray tube is placed inside a solenoid and electrons are accelerated towards the screen and deflected by a transverse AC voltage. 2. The resulting motion of the electrons is helical due to the magnetic field produced by the solenoid. By measuring the pitch of the helix, the e/m ratio can be calculated using the given formula. 3. The calculated value of e/m is 1.6 × 1011 C/kg with a percent error of 8.57% compared to the standard value of 1.75 × 1011 C/kg.

1. Dr. V. B. Bhatkar (HOD)
Dr. W. S. Barade
(BSc:- IIIrd
PCM, Sem - V)

2. Aim:- To determine specific charge “(e/m)” by Helical coil method.
Apparatus : - Cathode ray tube (CRT), CRT power supply, DC power supply (30V), solenoid,
Rheostat, DC voltmeter and connecting wires.
Formula:-
𝑒
𝑚
=
5×1013 ×𝐿2×𝑉𝑎
𝑁2 𝐼2 𝑙2
Where , l -- Distance to screen from deflecting plates
N -- Total number of turns of solenoid
L -- Length of the solenoid
V a -- Accelerating Voltage
I -- Solenoid current
Theory:- Electronics emitted from the cathode, accelerated by the anode are deflected by
electric field to give a line an the fluorescent screen.
The current carrying solenoid which enclose the cathode ray tube provides the necessary
magnetic field focusing it alternating potential is applied to the plates then the electrons shall
experience a transverse alternating force. Under in influence of this potential.
We shall get a line on the strength of the applied potential. Now if the longitudinal field because
of solenoid is electron in circular path is balanced by the centripetal forces supplied by the
magnetic field.
This is the popular method to find e/m. In this method to find e/m . In this method, the cathode
ray tube is placed inside a solenoid, if B is the magnetic filed to make the spot then and formula
used to calculate the of element.
𝑒
𝑚
=
5 × 1013
× 𝐿2
× 𝑉𝑎
𝑁2 𝐼2 𝑙2
Where , l -- Distance to screen from deflecting plates
N -- Total number of turns of solenoid
L -- Length of the solenoid

3. V a -- Accelerating Voltage
I -- Solenoid current
Basic methodology :- Electrons are accelerated towards the screen of a CRT and also
deflected by a transverse AC voltage. The CRT is placed in a magnetic field produced by a
solenoid. The resulting motion of the electrons in then helical. A measure of the pitch of the
helix leads to a calculation of the e/m ratio.
Figure:-
CRT
Procedure:
1. Record the constants of the solenoid and tube.
2. Place the solenoid such that its axis lies in the east west direction. Mount the cathode ray
tube inside the solenoid at the center. The power unit should be kept as far away as
possible to avoid the stray magnetic field.
3. Switch on the power supply unit and set the knob marked “Accelerating Voltage” and
adjust the voltage, V, to any desired value.

4. 4. Apply a.c deflecting potential to one set of plates, say X-plates. A deflection of 2cm is
adequate for the experiment.
5. Now turn on the solenoid current and increase the current till the line is reduced to a
small point. This current in amperes is Ix.
6. Repeat procedure of point 5 above with Y-plates. Keep deflection 2cm. This current in
amperes is I y .
7. Now repeat the whole procedure from point 4 to 5 with three other values of
accelerating voltages. It will be necessary to refocus the spot in the spot in the tube at
each voltage.
Observation:-
1. Distance between the edge of X-plate and the screen lx = 0.07 m.
2. Distance between the edge of Y-plate and the screen ly = 0.075 m.
3. Length of the winding L, = 0.2 m.
4. Number of turns, N = 1200 , D= 8.5 cm.
Sr.No. Accelerating Voltage
(Va) Volts
Solenoid current for x-
deflecting plates
Ix (Amp)
Solenoid current for y-
deflecting plates
Iy (Amp)
1. 645 1.03 0..93
2. 654 1.2 1.05
Calculation:-
𝑒
𝑚
=
5×1013 ×𝐿2×𝑉𝑎
𝑁2 𝐼2 𝑙2
1) For Va = 645V , Ix = 1.03 Amp , Iy = 0.93 Amp
a) lx= 0.07m
𝑒
𝑚
=
5 ×1013×(0.2)2×645
(1200)2(1.03)2(0.07)2
=
129×1013
7486.7
= 1.723 X1011
c/kg
b) ly =0.075m
𝑒
𝑚
=
5 ×1013×(0.2)2×645
(1200)2(0.076)2(1.03)2
=
129×1013
70005.69
= 1.84 X 1011
c/kg

5. 2) For Va = 654V , Ix = 1.2 Amp , Iy = 1.05 Amp
a) lx = 0.07m
𝑒
𝑚
=
5 ×1013×(0.2)2×645
(1200)2(0.07)2(1.2)2
=
130.8×1013
10160.64
= 1.28 X 1011
c/kg
b) ly = 0.075
𝑒
𝑚
=
5 ×1013×(0.2)2×654
(1200)2(0.075)2(1.05)2
=
130.8×1013
8930.25
= 1.464 X 1011
c/kg
Mean, (e/m) =
1.723 × 1011+1.84×1011+1.28×1011+1.464×1011
4
c/kg
= (
6.307
4
) × 1011 𝑐
𝑘𝑔
= 1.576 X 1011 ≈ 𝟏. 𝟔 × 𝟏𝟎 𝟏𝟏
𝒄/𝒌𝒈
% error =
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑
𝑒
𝑚
− 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑
𝑒
𝑚
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒/𝑚
× 100 %
=
(1.75−1.6)×1011
1.75 ×1011
X 100%
= 8.75 %
Results:- Calculated value of specific charge of electron (e/m) = 1.6 X 1011 C/Kg
Standard value of (e/m) = 1.75 X1011 C/Kg
% error in e/m = 8.57 %
Precautions:- 1. Solenoid should be at accurately placed with its axis in the east-west
direction.
2. CRT should be at the middle part of the solenoid.
3. Do not pass high currents through the solenoid for longer than necessary.

6. Viva – Question
1. What is Helical Coil Method?
Ans: In helical l method a cathode ray tube is inserted in a solenoid and e/m
determine the condition for focusing deflected electron to a spot on the screen.
2. What is the standard value of e/m?
Ans: The standard value of e/m is 1.75 X 1011.
3. SI unit of specific charge?
Ans: C/Kg
4. In which formula calculate the % error e/m?
Ans:% errorin e/m =
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑
𝑒
𝑚
− 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑
𝑒
𝑚
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒/𝑚
*Those who try, they fly …….*