print "Weekly Homework 9\n"
pi = 4*a(1)
epsilon = 8.85 * 10^(-12)
epercoul = 6.24 * 10^18
# Conductivity constants ( resistitvity = 1/sigma)
sigmairon = 1*10^7
sigmacopper = 6*10^7
sigmasilver = 6.2*10^7
sigmagold = 4.1*10^7
sigmaaluminum = 3.5*10^7
sigmatungsten = 1.8*10^7
sigmanichrome = 6.7*10^5
sigmacarbon = 2.9*10^4

print "Question 1\n"
#params
i = 1.9/1000 # A
diam = 5/1000 # m

r = diam/2
"Protons Per Second "
i*epercoul
# i = \int_0^R J_edge (r/R) 2\pi r dr
#   = J_edge/R [ 2\pi r^3 / 3 ]_0^R
#   = 2\pi J_edge R^2/3
jedge = 3*i/(2*pi*r^2)
"J_edge (A/m^2) "
jedge

print "Question 2\n"
# Equilibrium: eta/2epsilon + EF_I2 = EF_I1
# EF_I = V/L = IR/L = I(L/sigma/A)/L = I/(sigma*A)
# eta = 2epsilon * ( I/(sigma1*A)-I/(sigma2*A) )
# Corrections from pearson: apparently it should be eta/epsilon, and the
# sign was opposite, so correct is
print "eta = epsilon * ( I/(sigma2*A)-I/(sigma1*A))\n"
#params
diam = 1/1000 # m
i = 3 # A

a = pi*r^2 # m^2
r = diam/2 # m
"Charge on the boundary "
# eta * a
epsilon * i * (1/(sigmairon*a) - 1/(sigmacopper*a)) * a

print "Question 3\n"
#params
diam = .5 # mm
i = 20 # mA

diam = diam/1000 # m
r = diam/2 # m
i = i/1000 # A

a = pi*r^2
j = i/a
# j = ef*sigma
ef = j/sigmasilver
"Electric Field (V/m) "
ef

"Electron drift speed (m/s) " # idk lol
ve = 5.8 * 10^28 # number/m^3
ie = i*epercoul # number/s
ie/a/ve

print "Question 4\n"
diam = 1 # mm
diam = diam/1000 # m
r = diam/2
a = pi*r^2
# efnichrome = efaluminum
# jnichrome/sigmanichrome = jaluminum/sigmaaluminum
# i/pi*(diam_nichrome/2)^2/sigmanichrome = i/a/sigmaaluminum
# pi*(diam_nichrome/2)^2*sigmanichrome = a*sigmaaluminum
"Diameter of nichrome (m) "
2*sqrt(a*sigmaaluminum/sigmanichrome/pi)

print "Question 5\n"
#params
l = 10 # cm
innerdiam = 2.8 # mm
outerdiam = 3   # mm
v = 3 # V

l = l/100 # m
innerdiam = innerdiam/1000 # m
innerr = innerdiam/2
outerdiam = outerdiam/1000 # m
outerr = outerdiam/2
a = pi*outerr^2 - pi*innerr^2
ef = v/l
j = ef*sigmanichrome
i = j*a
"Current (A) "
i

print "Question 6\n"
print "60W means higher resistance, so it has a higher voltage drop\n"
print "and takes more energy\n"

v = 120 # V
w1 = 60 # W (wattage of lightbulb 1 @ 120V)
# w1 = i*v = v*v/r1
r1 = v^2/w1
r1
w2 = 100 # W
r2 = v^2/w2
i = v/(r1+r2)
print "Power by the first bulb (W) "
i^2*r1
print "Power by the second bulb (W) "
i^2*r2

print "Question 7\n"
#params
r = 70 # Ohm

v = 9 # V
i = v/r

"Magnitude of Current (A) "
i

print "Direction is left to right (positive to negative)\n"

print "Question 8\n"
#params
w = 1100 # W

v = 120 # V
"Resistance (Ohm) "
r = v^2/w
r

"Current (A) "
v/r

print "Question 9\n"
#params
m = 1 # g aluminum
p = 8 # W
v = 1.5 # V

m = m/1000 # kg
r = v^2/p # Ohm
densityaluminum = 2.7 # g/cm^3
densityaluminum = densityaluminum*1000 #kg/m^3
vol = m/densityaluminum # m^3
rhoaluminum = 1/sigmaaluminum # Ohm*m
# R = rhoaluminum * l / A = rhoaluminum * l / (vol/l) =
# rhoaluminum*l^2 / vol
l = sqrt(r/rhoaluminum*vol) # m
# Cross-sectional area
a = vol/l # m^2
"Diameter (mm) "
2*sqrt(a/pi) * 1000

"Length (m) "
l