a) Let's assume a nuclear reactor was a completely isolated system, such that no matter particles can escape from it and the only thing that can get out is energy in the form of heat and radiation. If we weigh the whole reactor before switching it on, and again after it has been operating for a while, and find that the mass difference is 0.05 grams, what is the total amount of energy that the reactor has produced?
b)Nuclear reactions can produce electrically charged particles (of say mass m and charge q). To measure the kinetic energy E of such a particle we want to find its speed v, and one way to do this in practice is to use a magnetic field (of constant magnitude B, and perpendicular to the particle's direction of motion). The Lorentz force law tells us that such a particle traveling though the magnetic field will experience a force of magnitude q v B. What direction does the force point in? Explain why this causes the particle to move in a circle, and derive a formula that gives the kinetic energy of the particle in terms of the (observable) radius of this circle.
2 answers
this sso guy is nuts if he thinks we should be able to do this...
a)
E = m c^2
here
m = .05 grams = .05*10^-3 kg = 5*10^-5 kg
c = 3*10^8m/s so c^2 = 9*10^16 m^2/s^2
so
E = 45 * 10^11 = 4.5*10^12 Joules
b) Force perpendicular to velocity gives centripetal acceleration
F = m v^2/R
so
q v B = m v^2/R
so
v = (q/m) B R
Ke = (1/2) m v^2
Ke = (1/2) m (q/m)^2 B^2 R^2
Ke = (1/2) (q B R)^2 sqrt(m)
E = m c^2
here
m = .05 grams = .05*10^-3 kg = 5*10^-5 kg
c = 3*10^8m/s so c^2 = 9*10^16 m^2/s^2
so
E = 45 * 10^11 = 4.5*10^12 Joules
b) Force perpendicular to velocity gives centripetal acceleration
F = m v^2/R
so
q v B = m v^2/R
so
v = (q/m) B R
Ke = (1/2) m v^2
Ke = (1/2) m (q/m)^2 B^2 R^2
Ke = (1/2) (q B R)^2 sqrt(m)