PY212 SPRING 2001 EXAM 2 SOLUTIONS

P1A: C = 2.17 mF. Q(t) = Q0 (1 - exp(-t/RC))

P1B: Measure current with AMMETER in SERIES. Ideal AMMETER has 
ZERO internal resistance.

P1C: mu = NIA = 0.1296 A m^2
     torque = mu x B = 2.84 x 10-2 N*m

P1D: Ampere's Law: Integral B*dA = mu0 I_enclosed... you can
     break up the closed path of Ampere's Law into several
     path segments.

     least = d = b (=0 because the path is perpendicular to B)
     c    = 1/4 mu0 I... work it out from B = mu0 I / 2 pi R
     e    = 3/4 muo I
     most = a = f (= mu0 I, because path is closed)

P2: Ohm's Law and Power
    (b) 20 Ohms
    (c) 1.2 A
    (d) 0.8 A ... this one takes the most work. For example, you can
    calculate calculate V across the parallel combination by I/R=4 Ohms.
    You calculated 4 Ohms to get part (b). I = 4.8 V / 6 Ohms
    (e) D ... use P = I^2 R. D sees all 1.2 A and has the largest R!
    (f) Bulb D gets dimmer. Removing bulb B increases the total resistance
    (to 22 Ohms). Now I = 24V/22 Ohms is small and I^2 R for bulb D is smaller.

P3: Superposition using B = mu0 I / 2 pi R
    (a) B = (mu0 I / 2 pi A) * sqrt(3) in the x-direction
    (b) x=0, y = -A/sqrt(3)
    (c) I = 500 A
    (d) the force between parallel currents is attractive
 
P4: (a) B = 1.1 x 10-3 T (use qvB = mv^2/r)
    (b) 181 turns
    (c) You need 113.75 meters of wire, R = rho l / A = 4 Ohms,
        radius = 0.5 mm
    (d) down (electrons are negative, field from current loop is in)