A free response practice
question on magnetic force on moving charges was posted on 8th July
2013. As promised I give below a model answer for your benefit, along with the
question:
The adjoining figure represents a cathode ray tube in a cathode ray
oscilloscope. The voltage applied (with respect to the cathode) to the final
accelerating anode is V and the distance
between the anode and the fluorescent screen of the cathode ray tube is L. The mass of the electron is m and the magnitude of its charge is e.
The cathode ray tube is so oriented that in the undeflected condition
the electron beam is horizontal and
is along the magnetic meridian, with
the electrons traveling from south to
north. The horizontal component of the earth’s magnetic field at the place
is Bh. Now, answer the
following questions:
(a) Calculate the velocity
with which the electrons strike the fluorescent screen, assuming that the space
between the accelerating anode and the fluorescent screen is devoid of electric
fields and the electrons start from the cathode with negligible speed.
(b) The cathode ray tube is rotated through 90º about a verti8cal axis so that the electrons travel from east to west.
In what direction will the spot of light (produced by the impact of the
electron beam) on the fluorescent screen be deflected? Select the correct
option from the following, by putting a tick mark (√) against one of the following:
Upwards………
Downwards………
Leftwards………
Rightwards……..
Justify your answer.
(c) If the cathode ray tube is rotated through 90º in the opposite direction, will the direction of deflection of the spot of
light on the screen change? Justify your answer.
(d) Derive an expression for the specific charge (charge to mass ratio)
of the electron in terms of the given data and the shift s of the electron spot on the screen, on rotating the cathode ray
tube through 90º. Assume that the shift s is
small compared to the distance between the
accelerating anode and the fluorescent screen.
Answer:
(a) The electrons acquire kinetic energy under the action of the
accelerating potential V applied on
the accelerating anode. If the final velocity acquired by the electrons is v, we have
½ mv2 = eV
Therefore v = √(2eV/m)
This is the velocity with which the electrons
strike the fluorescent screen since their velocity after leaving the aperture
of the accelerating anode remains unchanged in the field free space between the
anode and the screen.
(b) Upwards.
Earth’s magnetic field is directed from south to north. Electrons traveling from
east to west make a conventional current flowing from west to east (since the
electrons are negatively charged).
[The current produced by electrons traveling from
east to west has the same direction as the current produced by positive charges
traveling from west to east]
Therefore, in order to apply Fleming’s left hand
rule for fining the direction of the magnetic force on the electron beam, we
have to hold the forefinger along the south to north direction and the middle
finger along the west to east direction. The thumb will then point upwards.
(c) If the cathode ray
tube is rotated through 90º in the opposite direction,
the electron beam will be deflected downwards
since the electrons travel from west to
east.
(d) When the
electrons travel parallel to the horizontal component (Bh) of earth’s magnetic field, there is no
magnetic force on them. This follows from the expression for magnetic force F:
F
=evBh sinθ
where θ is the angle between the
magnetic field and the direction of the conventional current. Since θ = 180º, F
= 0.
When the cathode
ray tube is rotated through 90º, the electrons travel at right angles to the horizontal
component (Bh) of earth’s magnetic field. In this case the
magnetic force is maximum (Fmax)
and is given by
Fmax
= evBh.
Since the
deflection (s) produced is small (Fig.), we may assume that this force acts
vertically throughout the path of the electrons.
The above vertical
force produces a vertical acceleration and the vertical displacement s suffered by the electrons is given by
the equation of uniformly accelerated motion,
s
= ut + ½ at2 ……….. (i)
where u is the initial vertical velocity (at the instant of leaving the
accelerating anode), which is zero, t is the time taken by the electron to
travel from the accelerating anode to the fluorescent screen and a is the vertical acceleration which is
given by
a
= Fmax/m = evBh/m
The time taken by the electron to travel the
distance L between the anode and
screen is given by
t = L/v
Substituting for u, a and t in Eq. (i) we have
s = 0 + ½ (evBh/m) (L/v )2 = ½ (e/m) (BhL2/v)
The speed v of the electrons, as shown in part (a)
is √(2eV/m)
Substituting for v we have
s = ½ (e/m) (BhL2/√(2eV/m)
This gives e/m = 8s2V/ Bh2L4
