Page 351 - Schaum's Outline of Theory and Problems of Applied Physics
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336                           ELECTROMAGNETIC INDUCTION                          [CHAP. 28



                                                  B





                                         N turns




                                                        Cross-sectional area A
                                                 Fig. 28-2

        symbol   (Greek capital letter phi):

                                          = BA
                               Magnetic flux = (magnetic field)(cross-sectional area)

                                                             2
        The unit of magnetic flux is the weber (Wb), where 1 Wb = 1T·m . According to Faraday’s law of electromag-
        netic induction, the emf induced in the coil when the magnetic flux through it changes by    in the time  t is
        given by

                                        Induced emf = V =−N
                                                               t



        LENZ’S LAW

        The minus sign in Faraday’s law is a consequence of Lenz’s law: An induced current is always in such a direction
        that its own magnetic field acts to oppose the effect that created it. For example, if B is decreasing in magnitude in
        the situation of Fig. 28-2, the induced current in the coil will be counterclockwise in order that its own magnetic
        field will add to B and so reduce the rate at which B is decreasing. Similarly, if B is increasing, the induced
        current in the coil will be clockwise so that its own magnetic field will subtract from B and thus reduce the rate
        at which B is increasing.


        SOLVED PROBLEM 28.1
              The vertical component of the earth’s magnetic field in a certain region is 3×10 −5  T. What is the potential
              difference between the rear wheels of a car, which are 1.5 m apart, when the car’s velocity is 20 m/s?
                  The rear axle of the car may be considered as a rod 1.5 m long moving perpendicular to the magnetic field’s
              vertical component. The potential difference between the wheels is therefore

                               V = Blv = (3 × 10 −5  T)(1.5m)(20 m/s) = 9 × 10 −4  V = 0.9mV

        SOLVED PROBLEM 28.2

              A square wire loop 8 cm on a side is perpendicular to a magnetic field of 5 × 10 −3  T. (a) What is the
              magnetic flux through the loop? (b) If the field drops to 0 in 0.1 s, what average emf is induced in the
              loop during this time?
              (a) The area of the loop is
                                           (8cm)(8cm)         2         −3  2
                                       A =            = 0.0064 m = 6.4 × 10  m
                                           (100 cm/m) 2
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