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Chapter22

VIEWS: 45 PAGES: 7

									Chapter 22 – Conceptual Questions
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1.   REASONING AND SOLUTION If the coil and the magnet in Figure 22.1a were each
     moving with the same velocity relative to the earth, there would be no relative motion
     between the magnet and the coil. The magnetic flux through the coil due to the bar magnet
     would be constant and, therefore, the combined motion of the bar magnet and the coil would
     not result in an induced current in the coil. We are ignoring here any effect due to the earth's
     magnetic field.
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2.      REASONING AND SOLUTION In the discussion concerning Figure 22.5, we saw that
a force of 0.086 N from an external agent was required to keep the rod moving at constant speed.
Suppose the light bulb in the figure is unscrewed from its socket. Once the light bulb is
unscrewed, the conducting rails and the rod are no longer part of a complete circuit (the
resistance of the empty socket is infinite). Therefore, even though there will be a charge
separation and a motional emf in the rod given by vBL, there will be no current in the rod. Since
there is no current, there is no magnetic force to resist the motion of the rod. From Newton's first
law, the rod will continue to move with constant velocity v without the application of an external
force.

4.   REASONING AND SOLUTION A magnetic field B is necessary if there is to be a
     magnetic fluxpassing through a coil of wire. Yet, just because there is a magnetic field
     does not mean that a magnetic flux will pass through a coil.
     The general expression for magnetic flux is given by Equation 22.2:   (B cos )A , where
     B is the magnitude of the magnetic field, A is the cross-sectional area of the coil, andis the
     angle between the magnetic field B and the normal to the surface of the coil. Equation 22.2
     shows that the flux depends only on the component of the magnetic field that is
     perpendicular to the surface of the coil. As shown in Example 4 and in Figure 22.11, when
     the coil is oriented so that it is parallel to the field,  = 90°, B has no component normal to
     the surface of the coil, and the magnetic flux through the coil is zero. Therefore, the
     magnetic flux through a coil can be zero even though there is a magnetic field present.


6.   REASONING AND SOLUTION Initially, before the switch is closed, neither the con-
     ducting rails nor the rod carries a current. When the switch is closed, a conventional current
     will flow along the conducting rails from the positive toward the negative terminal of the
     battery. Since the rod is a conducting rod, current will flow through the rod, from top to
     bottom. According to RHR-1, there will be a force that points to the right on the conducting
     rod due to the magnetic field; therefore, the rod will be pushed and accelerate to the right.
     As the rod moves to the right, the area bound by the "loop" increases, thereby increasing the
     magnetic flux through the loop. As the magnetic flux increases, an induced emf appears
     around the "loop." According to Lenz's law, the induced emf that appears, will appear in
     such a way so as to oppose the increase in the magnetic flux. This will occur if the induced
     emf opposes the battery emf, with the result that the current in the rod begins to decrease
     and reaches zero when the induced emf exactly offsets the battery emf. With no current in
     the rod, there is no longer a magnetic force applied to the rod. With no force, there is zero
     acceleration. In other words, from this point on, the rod moves with a constant velocity.
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7.   REASONING AND SOLUTION A bolt of lightning contains moving charges, and hence,
     is a current. This current is surrounded by its magnetic field. Since the charges in a bolt of
     lightning move erratically, the current is a time-dependent and gives rise to a time-
     dependent magnetic field. Many household appliances contain coils. If the time-dependent
     magnetic field passes through these coils, the magnetic flux through the coils will be time-
     dependent. From Faraday's law, there will be an induced emf and, hence, an induced current
     in the coils of the appliance. This will result in an induced current in the circuit that contains
     the appliance. Therefore, a bolt of lightning can produce a current in the circuit of an
     electrical appliance, even when the lightning does not directly strike the appliance.
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12. REASONING AND SOLUTION                    The                                Met al ring
    figure shows the set-up.                           I ron core

     a. When the switch is open, there is no                                                  Switch
     current in the coil and no magnetic field is
     present. When the switch in the circuit is
     closed, a current is established in the coil.            Coil
     From RHR-2, the magnetic field asso-                                                  +    –
     ciated with the current in the coil leaves
     the bottom of the coil and enters the top                                            Bat tery
     of the coil (see drawing below). In other
     words, when there is a current in the coil,
     the bottom of the coil acts like the north
     pole of a bar magnet, and the top of the coil acts like the south pole of a bar magnet. The
     magnetic field of the coil causes a magnetic flux through the metal ring.
         In the very short time that it takes for the current to rise from zero to its steady value, the
     magnetic field associated with the current also rises from zero to its steady value. As the
     magnetic field increases, the magnetic flux through the metal ring changes, and there is an
     induced voltage and an induced current around the ring.
         From Lenz's law, the polarity of the
     induced voltage will be such that it                                   N
     opposes the changing flux through the                                                  Induced
     metal ring. The induced magnetic field of                                          magnetic field
                                                                            S
     the metal ring will be directed upward
     through the ring to subtract from the                                  S             I
     magnetic field of the coil that points                                         Coil
     downward through the ring. This is
     illustrated in the figure at the right. To                                         +     –
     preserve clarity, the iron core and a                                   N
     portion of the circuit have not been                                             I
                                                       Magnetic field
     shown. The induced magnetic field of the           due to coil
     metal ring is similar to that of a bar
     magnet, with the north pole above the
     ring, and the south pole below the ring.
     Thus, the situation is similar to that of two
     bar magnets with their south poles next to
     each other. The two south poles repel,
     and the ring "jumps" upward.
    b. If the polarity of the battery were reversed, the ring still "jumps" upward when the switch
    is closed. When the polarity of the battery is reversed, and the switch is closed, the current
    moves through the coil in the opposite direction.
    The magnetic field associated with the
    current points in a direction that is
                                                                          S
    opposite to the direction of the magnetic                                          Induced
    field in part (a). In other words, the top of                                     magnetic field
    the coil acts like the north pole of a bar Magnetic field              N
    magnet, and the bottom of the coil acts           due to coil
    like the south pole of a bar magnet.                                    N           I
       When the switch is closed, the                                            Coil
    magnetic flux through the metal ring                                          I –
    changes, and there is an induced voltage                                             +
    and an induced current around the ring.                                 S

    In accord with Lenz's law, the induced magnetic field of the metal ring will be directed
    downward through the ring to subtract from the magnetic field of the coil that points upward
    through the ring. The induced magnetic field of the metal ring is similar to that of a bar
    magnet, with the south pole above the ring, and the north pole below the ring. The two
    north poles repel, and the ring "jumps" upward.
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Chapter 22, Problems

22-2. REASONING AND SOLUTION
        a.       The right-hand rule shows that positive charge would be forced to the
 driver's side .

       b.      According to Equation 22.1, the width of the car is

                                                    –3
                             Emf         2.4  10 V
                          L                               2.0 m
                              vB                
                                    25 m/s  4.8  10–5 T    
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22-3. REASONING AND SOLUTION Using Equation 22.1, we find

                Emf = vBL = (220 m/s)(5.0  10–6 T)(59 m) = 0.065 V
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22-4. REASONING AND SOLUTION The motional emf generated by a conductor moving
perpendicular to a magnetic field is given by Equation 22.1 as Emf = vBL, where v and L are the
speed and length, respectively, of the conductor, and B is the magnitude of the magnetic field.
The emf would have been

                                                                  
               Emf  vBL  7.6  103 m/s 5.1  105 T 2.0  104 m  7800 V
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22-5. WWW REASONING AND SOLUTION For each of the three rods in the drawing,
    we have the following:

    Rod A: The motional emf is zero , because the velocity of the rod is parallel to the
    direction of the magnetic field, and the charges do not experience a magnetic force.

       Rod B: The motional emf is, according to Equation 22.1,

                          Emf = vBL  (2.7 m/s)(0.45 T)(1.3 m) = 1.6 V

       The positive end of Rod 2 is end 2 .


       Rod C: The motional emf is zero , because the magnetic force F on each charge is
directed perpendicular to the length of the rod. For the ends of the rod to become charged, the
magnetic force must be directed parallel to the length of the rod.
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22-10. REASONING AND SOLUTION
       a.   The magnetic flux is  = BA cos . Since  = 0.0°, we have

                         = (0.35 T)(0.0160 m2) cos 0.0° = 5.6  10 –3 Wb

       b.         Since  = 90°, we have

                      = (0.35 T)(0.0160 m2) cos 90° = 0 Wb
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22-11. SSM REASONING The general expression for the magnetic flux through an area A
is given by Equation 22.2:   BAcos where B is the magnitude of the magnetic field and 
is the angle of inclination of the magnetic field B with respect to the normal to the area.

       The magnetic flux through the door is a maximum when the magnetic field lines are
                                                                      )
perpendicular to the door and 1  0.0 so that 1  max  BA(cos 0.0 = BA.

       SOLUTION When the door rotates through an angle 2 , the magnetic flux that passes
through the door decreases from its maximum value to one-third of its maximum value.
                1
Therefore  2  3 max , and we have

             2  BA cos 2  1 BA
                              3
                                        or      cos 2      1
                                                             3
                                                                     or       2  cos –1     1
                                                                                                 3
                                                                                                       70.5
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22-12. REASONING The magnetic flux is defined in Equation 22.2 as   BA cos  , where B
is the magnitude of the magnetic field, A is the area of the loop, and  is the angle between the
normal to the surface of the loop and the magnetic field. The change  in flux is the final flux
 minus the initial flux 0;  =  – 0.

SOLUTION Let L1 and L2 be the lengths of the left and bottom edges of the loop. The initial
area is the area of this rectangular loop plus the area of the semicircle: A0 = L1  L2 + 1  r 2 . The
                                                                                          2

                                                  
initial magnetic flux is 0 = B L1  L2 + 1  r cos 0 . The final area is the area of the
                                          2
                                               2



rectangular loop minus the area of the semicircle: A0 = L1  L2  1  r 2 . The final magnetic flux
                                                                  2

                               
is  = B L1  L2  1  r cos 0 . The change in flux is
                   2
                        2




                                    
                 =    0  B L1  L2  1  r
                                           2
                                                   2
                                                        cos 0°  B  L  L
                                                                      1   2
                                                                                1 r
                                                                                 2
                                                                                        2
                                                                                             cos 0°
                    =  r cos 0     T    0.20 m  cos 0  9.4  10
                            2                                    2                          2
                                                                                                 Wb

    Note that the change in flux does not depend on L1 or L2.
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22-17. REASONING AND SOLUTION
       a.      The change in flux through the loop is  = (B)A cos , so the magnitude for
the emf given by Faraday's law is

                                       Emf = N[(B)/(t)]A cos 

                Emf = 300[(0.40 T)/(0.80 s)](5.0  10–3 m2) cos 30.0° = 0.65 V

       b.      The current in the resistor is, according to Ohm's law,

                      I = (Emf)/R = (0.65 V)/(6.0 ) = 0.11 A
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22-18. REASONING AND SOLUTION The average emf induced in the circular wire is given
    by

                                                   A  A0     
                                        Emf   B 
                                                   t t        cos 
                                                               
                                                        0     

       The change in the area is equal to the final area of the circle (A = 0 m2) minus the initial
area (A0 =  r2).
                           0   r2    
                Emf   B 
                           t t         cos 
                                        
                                 0     


            Emf  (0.55 T) 
                                            
                             0   2.0  102 m 2 
                                                           
                                                    cos 0.0  2.8  103 V
                                   0.25 s  0     
                            
                                                  
                                                   
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22-19. SSM REASONING According to Equation 22.3, the average emf induced in a coil
of N loops is Emf = –N  /  t .
       SOLUTION For the circular coil in question, the flux through a single turn changes by

                               BA cos 45 – BA cos 90  BA cos 45

      during the interval of t  0.010 s. Therefore, for N turns, Faraday's law gives the
magnitude of the emf (without the minus sign) as

                                                        NBA cos 45
                                                Emf =
                                                            t

       Since the loops are circular, the area A of each loop is equal to  r2. Solving for B, we
have

                            Emf  t
                                   (0.065 V)(0.010 s)
                   B                                    8.6  10 –5 T
                   N r cos 45 (950) (0.060 m) cos 45
                             2                  2


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22-20. REASONING According to Faraday’s law as given in Equation 22.3 (without the minus
sign), the magnitude of the emf is Emf = /t for a single turn (N = 1). Since the normal is
parallel to the magnetic field, the angle  between the normal and the field is  = 0° when
calculating the flux  from Equation 22.2:  = BA cos 0° = BA. We will use this expression for
the flux in Faraday’s law.

       SOLUTION Representing the flux as  = BA, we find that Faraday’s law (without the
minus sign) becomes
                                        BA B  A
                            Emf =                
                                     t      t       t

        In this result we have used the fact that the field magnitude B is constant. Rearranging
this equation gives
                               A Emf 2.6 V
                                                   1.5 m 2 /s
                              t      B      1.7 T
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22-21. REASONING AND SOLUTION Faraday's law gives (using only the magnitude of the
induced emf)
             t = BA cos 0.0°/(Emf) = (1.5 T)(0.032 m2)/(0.010 V) = 4.8 s
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22-35. SSM WWW REASONING AND SOLUTION
        a. From the drawing, we see that the period of the generator (the time for one full cycle)
is 0.42 s; therefore, the frequency of the generator is

                                               1   1
                                         f            = 2.4 Hz
                                               T 0.42 s

       b. The angular speed of the generator is related to its frequency by   2  f , so the
angular speed is
                                      2 (2.4 Hz) = 15 rad/s

        c. The magnitude of the maximum emf induced in a rotating planar coil is given by
 Emf 0  NAB (see Equation 22.4). The magnitude of the magnetic field can be found by
solving this expression for B and noting from the drawing that (Emf)0 = 28 V:

                               Emf 0    28 V
                         B                               0.62 T
                        NA                   2
                                (150)(0.020 m )(15 rad/s)
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