PHYSICS 220BL FINAL EXAMINATION SPRING 2001

1. Three long parallel wires are 40.0cm from one another. (Looking along them they are at three corners of an equilateral triangle.) the current in each wire is 8.00A,but that in wire M is opposite to that in wires N and P (Figure 1). Determine the magnetic force per unit length on M due to wires N and P.

Solution Because the currents and the separations are the same, we find the force per unit length between any two wires from

                      F/L = I(m0I2/2pd) = m0I2/2pd

                            = (4p x 10-7T.m/A)(8.00A)2/2p(0.400m)

                            = 3.2 x 10-5N/m.

             The directions of the forces are shown on the diagram. We have

                                   FM/L = 2(F/L)cos30°

                                = 2(3.2 x 10-5N/m)cos30° = 5.54 x 10-5N/m up.

                     

  

                        Fig. 1

2. In Figure 2 the rod moves with a speed of 1.8m/s, is 24.0cm long, and has 2.2W resistance. The magnetic field is 0.35T and the resistance of the U-shaped conductor is 26.0W at a given instant. Calculate: (a) the emf induced; (b) the current flowing in the U-shaped conductor, and (c) the external force needed to keep the rod's velocity constant at that instant.

Solution. (a) Because the velocity is perpendicular to the magnetic field and the rod, we find the induced emf from

                         e = Blv 

                            = (0.35T)(0.240m)(1.8m/s) = 0.15V.

              (b) We find the induced current from

                         I = e/R = (0.15V)/(2.2W + 26.0W) = 5.4 x 10-3A

              (c) The induced current in the rod will be down. Because this current is in an outward magnetic field, there will be a magnetic force to the left. To keep the rod moving, there must be an equal external force to the right, which we find from

                        F = IlB = (5.4 x 10A)(0.240m)(0.35T) = 4.5 x 10-4N.

  

3. Two solenoids A and B, spaced close to each other and sharing the same cylindrical axis, have 400 and 700 turns, respectively. a current of 3.5A in coil A produces a flux of 300mWb at the center of B. (a) Calculate the mutual inductance of the two solenoids. (b) What is the self-inductance of A? (c) What emf is induced in B when the current in A increases at the rate of 0.50A/s?

    Solution.

         (a) M21 = (N2F21)/I = 700(90 x 10-6Wb)/3.5A = 18mH.

         (b) L = NF /I = 400(300 x 10-6Wb)/3.5A = 34mH.

         (c) e2 = -M21dI1/dt = -(18 x 10H)(0.50A/s) = -9.0mV.

4. A coil of resistance 3.50W and inductance 20.5H is in series with a capacitor and 200-V(rms), 100-Hz source. The rms current in the circuit is 4.0A. (a) Calculate the capacitance in the circuit. (b) what is Vmax across the coil? (c) Draw a phasor diagram for this circuit.

Solution.  We firsst calculate the impedance:

                Z = Vrms/Irms = 200V/4.00A = 50.0W = Ö R2 + (XL - XC)2

               (50.0W) = (35.0W) + [2p(100Hz)(20.5H) - XC)2

               +35.7W = 12880W -  XC

           (a) Either XC = 12920W = 1/2p(100Hz)C  and  C = 123nF

                or       XC = 12840W = 1/2p(100Hz)C  and  C = 124nF.

           (b) VL = IrmsXL = I2pfL = (4.00A)2p)(100Hz)(20.5V.s/A) = 51.5kV.

           (c)