Solutions to Week 6 Assignment

1. (a) The increasing current in the wire will cause an increasing field into the page through the loop. To oppose this increase, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise.     (b)  The decreasing current in the wire will cause a decreasing field into the page through the loop. To oppose this decrease, the induced current in the loop will produce a flux into the page, so the direction of the induced current will be clockwise.    (c) Because the current is constant, there will be no change in flux, so the induced current will be zero.    (d) The increasing current in the wire will cause and increasing field into the page through the loop. To oppose this increase, the induced current in the loop will produce a flux out of the page, so the direction of the induced current will be counterclockwise.

2. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from                      e = BLv                         = (0.800T)(0.120m)(0.150m/s)                         = 1.44 x 10-2V = 14.4mV.     (b) Because the upward flux is increasing, the induced flux will be into the page, so the induced current is clockwise. Thus the induced emf in the rod is down, which means that the electric field will be down. We find its magnitude from                     E = e/l = (1.44 x 10-2V)/(0.120m) = 0.120V/m down.

3. (a) For the resistance of the loop, we have

                    R = rL/A = (1.68 x 10^-8W·m)(20)p(0.310m)/(1.3 x 10^-3m)² = 0.0616W.

         The induced emf is

                    e  = -DF_B/Dt = -ADB/Dt = -(20)p(0.155m)²(8.65 x 10^-3T/s) = -0.0131V.

         Thus the induced current is

                    I = e/R = (0.0131V)/(0.0616W) = 0.21A.

    (b) Thermal energy is produced in the wire at the rate of

                   P = I²R = (0.21A)²(0.0616W) = 2.8mW.

4.  (a) Because V_S < V_P, this is a step-down transformer.

     (b) We assume 100% efficiency, so we find the current in the secondary from

                   P = I_SV_S;

                  40W = I_S(12V), which gives I_S = 3.3A.

    (c) We find the current in the primary from

                  P = I_PV_P;

                 40W = I_P(120V), which gives I_P = 0.33A.

    (d) We find the resistance of the bulb from

                 V_S = I_SR_S;

                 120V = (3.33A)R_S, which gives R_S = 3.6W.

5. (a) If we assume the closed approach of Mars to Earth, we have

                Dt = L/c = [(227.9 - 149.6) x 10⁹m]/(3.00 x 10⁸m/s) = 2.6 x 10²s = 43.min.

    (b)  If we assume the closed approach of Saturn to Earth, we have

                Dt = L/c = [(1427 - 149.6) x 10⁹m]/(3.00 x 10⁸m/s) = 4.3 x 10³s = 71.min.