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Albert Einstein

Thursday, July 11, 2013

AP Physics C – Magnetic Fields – Answer to Free Response Practice Question on Magnetic Force on Moving Charges

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:
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.
(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

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