ANSWER TO CHAPTER 29 HOMEWORK

1.  (a)  The average induced emf is

                         e = -DFB/Dt = -ADB/Dt = -p(0.10m)2[(-0.45T) - (+0.52T)]/(0.180s) =  0.17V.

         (b)  The positive reult for the induced emf means the induced field is away from the observer, so the induced current is clockwise.

2.  The flux through the loop is

                         F = AB = AmonIo cos wt.

         The induced emf is

                         e = -dFB/dt = -BdA/dt = -(0.48T)(-3.5 x 10m2/s) = 1.7 x 10-2V.

         Because the area changes at a constant rate, this is the induced emf for both times.

3.  (a) Although there is an induced emf, there is no current because there is no closed circuit. Thus the rod

                will move at constant speed.

         (b) When the circuit is completed, there will be a current.

               The induced emf is Blv. Because the only resistance is from the rod, the current is

                            I = Blv/R.

                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, which will produce an acceleration:

                            F = -IBl = -B2l2v/R = ma = mdv/dt, or -(B2l2/mR)dt = dv/v.

                We integate to get the speed:

                           ò0t (-B2l2/mR)dt = òvovdv/v;

                            -B2l2t/mR = ln(v/vo), or v = ve-B2l2t/mR.

                 The speed eventually goes to zero. This is an application of Lenz's law.

  

4.  We find the rotation speed from

                            epeak = NBAw;

                            120V = (420turns)(0.350T)(0.210m)2w, which gives w = 18.5rad/s =  2.95rev/s.

5.  (a) We assume 100% efficiency, so we find the input voltage from

                             P =IPVP;

                             100W = (26A)VP, which gives VP= 3.8V.

                Because VS > VP, This is a   step-up  transformer.

          (b) For the voltage ratio we have

                              VS/VP  = (12V)/(3.84V) =   3.1.

6.  The induced emf around a circle is

                             e = -dFB/dt = - AdB/dt.

         Because A and dB/dt are the same for the two circular paths, e is the same. 

         even though E is greater in the region of the outer circle, the integral

         ò E × dl is the same. Note that for part of the path the component of E is 

          parallel to dl and for part it is antiparallel.

 

  
7*. (a) At a distance r from the long wire, the magnetic field is directed into the paper with magnitude

                      B = moI/2pr.

Because the field is not constant over the short section, we find the induced emf by integration. We choose a differential element dr a distance r from the long wire.

The induced emf in this element is 

                     de = Bvdr toward the long wire.

We find the total emf by integrating:

                      e  = òBvdr = moIv/2pòdr/r 

                           = (moIv/2p)ln[(a+ b)/b] toward long wire.

(b) If the current is in the opposite direction to I, the only change will be in the direction of the emf:

                      e  = (moIv/2p)ln[(a+ b)/b] away from long wire.

 
8*. Without the transformers, we can find the delivered current, which is the current in the transmission lines, from the delivered power:
                        Pout =  IVout;

                        65 x 103W = I(120V), which gives I = 542A.

The power loss in the lines is

                        PL0 = I2Rlines = (542A)2(2)(0.100W) = 5.87 x 104W = 58.7kW.

With the transformers, to deliver the same power at 120V, the delivered current from the step-down transformer is still 542A.

If the step-down transformer is 99% efficient, we have

                        (0.99)Ip2Vp2 = Is2Vs2;

                        (0.99)Ip2(1200V) = (542A)(120V), which gives Ip2 = 54.7A.

Because this is the current in the lines, the power loss in the lines is

                        PL2 = Ip22Rlines = (54.7A)2(2)(0.100W) = 5.99 x 102W = 0.599kW.

For the 1% losses in the transformers, we approximate the power in each transformer.

                        PLt = (0.01)(65kW + 65.6kW) = 1.31kW.

The total power loss using the transformers is 

                         PL =  0.599kW + 1.31kW = 1.9kW.

The power saved by using the transformers is

                          Psaved = PL0 - PL = 58.7kW - 1.9kW = 56.8kW.

 

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