The device at right provides a beam of negatively charged particles (drawn as an
ID: 3278988 • Letter: T
Question
The device at right provides a beam of negatively charged particles (drawn as an orange line). Let us call the direction along the beam x, the "up" direction y, and the axis mostly toward as z. Your job is to design a device (to be placed where I put the xyz axes) that allows electrons (with mass m_e and charge magnitude e) to go straight through, while deflecting more massive particles toward larger z, and Cathode less massive particles toward smaller z, (a) In what direction should the electric field point? Remember that these are negatively charged particles! (b) In what direction should the magnetic field point? Remember that these are negatively charged particles! (c) What is the relationship between electric and magnetic field m of the voltage V rather than the speed v. (d) If a particle of arbitrary mass m and charge q leaves the cathode, what lateral acceleration does your part of the device impose on it? Please eliminate E from your final expression and simplify as much as possible. (e) If two particles have different q and different m but the same ratio q/m, can this device distinguish between them? Explain why or why not.Explanation / Answer
a. the device is placed where the coordinate axes are placed
so for the -vely charged particles, which ar eelectrons, they should pass through
heavier particles should be deflected towards +ve z axis and lighter particles towards -ve z
so, Assume we have an electric field along +z axis,
so the force on -vely charged particles is towards the -ve z axis
to compensate for this force, we have a magnetic field in -y direction, then the force on -vely charged particles will be along +z axis
hence electric field can point towards +z axis
b. Then the magnetic field should point towards -y axis
c. For electrons , these forces are equal in magnitude
so for charge of electrn, e and mass m
force due to electric field = eE
force due to magnetic field = evB [ where E and B are electric and magnetic field magnitudes and v is velocity of the electron]
now, potential difference provides the inetic energy to electrons
eV = 0.5mv^2
v = sqroot(2eV/m)
hence eE = evB
E = vB = B*sqroot(2eV/m)
E^2/B^2 = 2eV/m
d. for a particle of charge q, and mass m,
lateral acceleration = Force/mass = (qB*sqroot(2qV/m) - qE)/m
but E = B*sqroot(2eV/m)
a = (qB*sqroot(2qV/m) - q*B*sqroot(2eV/me))/m [ here me is mass of electron]
a = qB*sqroot(2V)[sqroot(q/m) - sqroot(e/me)]/m
e. no this device cannot differentiate between two different particles of same q/m ratio, becAUse FROM PREVIOUS FORMULA A = 0 IF Q/M IS THE SAME FOR PARTICLES