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+########################################################################
+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