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Physics Semester 2  STPM  Chapter 14 Electric Current

               Section  B   Structured Questions
              Question 5
                A current flows in a wire of length L and radius r with the free electrons travelling
                                            –1
                at a drift velocity of 6.25 × 10  m s  for 1.0 minute.
                                       –3
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                (a)  If the number of free electrons in 1.0 m  of wire is 5.0 × 10 , calculate the
                                                                  19
                                                   3
                    current flow in the circuit.
                (b)  Calculate the drift velocity for the same current in a wire of the same
                                        1
                    materials but with radius  r.
                                        2
                (c)  The wire is then replaced with a wire of the same length that tapers
                    uniformly from end P to end Q as shown in the diagram.
                                 P                             Q

                    The diameter at  P is twice that at  Q. Compare the drift velocity of the
                    conduction electrons at P and at Q.

              Answer:

                 (a)   N  = 5.0 × 10      N = 5.0 × 10 19
                               19
                     A
                     Total charge = Ne
                               = (5 × 10 )(1.6 × 10 )
                                       19
                                                –19
                               = 8.0 C
                                Q
                     Current, I  =
         Semester
         2                       t
                              =   8    = 0.13 A
                                1 × 60
                 (b)  v =   I   =   I
                         nAe   nπr e
                                  2
                            v ∝   1
                               r 2
                               =    2
                            v 2   r 1  2
                            v 1  1 r 2
                         v 2   =   r  2
                                  1
                     6.25 × 10 –3  1 2
                                   r
                                  2
                               = 4
                           v 2  = 4(6.25 × 10 )
                                        –3
                             = 0.025 m s –1


                                              224




         014 STPM Q&A Physics T2.indd   224                                  17/02/2022   10:12 AM
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