ANSWER TO CHAPTER 27 HOMEWORK

1.  (a)  We see from the diagram that the magnetic field is up, so the top pole face is a  south pole.

         (b) We find the current from the length of wire in field:

                       F = ILB;

                      5.30N = I(0.10m)(0.15T), which gives I =  3.5 x 102A.

         (c)  The new force is

                      F = ILBsinq = Fsinq = (5.30N)sin80° =  5.22N.

               Note that the wire could be tipped either way.

  

2.  (a) The magnetic force provides the centripetal acceleration:

                       qvB = mv2/r,  or mv = qBr.

               The kinetic energy of the electron is  

                      K = 1/2mv2 = 1/2(qBr)2/m = (q2B2/2m)r2.

          (b)  The magnetic force provides the centripetal acceleration:

                      qvB =mv2/r,  or  mv = p = qBr.

                 The angular momentum is

                      L = mvr = qBr2.

3.  The total force on the proton is

                     F = e(E + v x B)

                        = e{(3.0i -4.2j) x 103V/m + [(6.0i + 3.0j - 5.0k) x 103m/s x (00.45i + 0.20j)T]}

                        = (1.60 x 10-19C)[(3.0i - 4.2j) + (1.0i - 2.25j - 0.15k)] x 103V/m

                        =   (6.4i - 10.3j - 0.24k)] x 10-16N.

4.  (a)  The angle between the normal to the coil and the field is 24.0°, so the torque is

                    t = NIABsinq

                      = (12)(7.10A)p(0.0850m)2(5.50 x 10-5T)sin24.0°

                      =   4.33 x 10-5m.N.

         (b)  From the directions of the forces shown on the diagram, 

                 the    north edge   of the coil will rise.

  
5*.  If we select a differential length dl of the wire, we see that the force on this element has a magnitude

                       dF = IBdl

and will have a direction perpendicular to the wire at an angle q below the horizontal. For each element there will be one diametrically opposite with an opposite horizontal force component, so the net force will be vertical. We find the net force by adding the vertical components from all of the differential elements:

                       F = òdFz = òIBdl(-sinq )

                          = -IB sinq òdl

                          = -IB2prsinq 

                          =  - 2pIBr2/(r2 + d2)1/2  (downward).

6*. The rotating charge is equivalent to a circular current. We choose a differential element of length dy a distance y from the axis of rotation. The charge on this element is

                     dq = (Q/l)dy.

Because the time for one revolution is T = 2p/w, the effective current of the element is

                     dl = dq/T = (w/2p)dq.

Thus the magnetic moment of the element is

                     dm = Adl = (py2)(w/2p)dq = (wQ/2l)y2dy.

We find the total magnetic moment by adding (integrating) the magnetic moments of the differential elements:

                     m  = òdm  = (wQ/2l)ò0ly2dy

                         = (wQ/2l)(l3/3) = wQl2/6.

 

HOME  SEND QUESTIONS