Ch30Word - Chapter 30 p 1 CHAPTER 30 Inductance and...

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Chapter  30,  p. 1 CHAPTER 30 – Inductance; and Electromagnetic Oscillations 1. The magnetic field of the long solenoid  is essentially zero outside the solenoid.  Thus there will be the  same linkage of flux with the second  coil and  the mutual inductance will be the same: M  =  μ 0 N 1 N 2 A / ¬ . 2. ( a ) We find the mutual inductance from M  =  μ N 1 N 2 A / ¬   = (2000)(4p × 10 –7  T     m/A)(300 turns)(100 turns)p(0.0200 m) 2 /(2.44 m)  = 3.88 × 10 –2  H =       38.8 mH . ( b ) The induced  emf in the second  coil is å   = –  M  ? I 1 /? t = – (3.88 × 10 –2  H)(0 – 12.0 A)/(0.0980 s) =        4.75 V . 3. The magnetic field inside the outer solenoid  is B  =  μ 0 n 1 I 1 . The magnetic flux linked  with the inner solenoid  is Φ 21  =  μ 0 n 1 I 1 A 2 . Thus the mutual inductance with the  N 2  turns of the inner solenoid  is M 21  =  N 2 Φ 21 / I 1  =  n 2 ¬ μ 0 n 1 A 2   , so the mutual inductance per unit length  is M / ¬  =  μ 0 n 1 n 2 p r 2 2 . 4. We find the mutual inductance of the system  by finding  the mutual inductance of the small coil.  The  magnetic field inside the long solenoid  is along the axis with a magnitude B 1  =  μ 0 n 1 I 1  =  μ 0 N 1 I 1 / ¬ . We find the magnetic flux through  the coil by using the area perpendicular to the field: Φ 21  = ( μ 0 N 1 I 1 / ¬ ) A 2  sin  θ . Thus the mutual inductance with the  N 2  loops of the coil is M 21  =  N 2 Φ 21 / I 1  =        ( μ 0 N 1 N 2 A 2 / ¬ ) sin  θ . 5. We find the mutual  inductance of the system  by finding  the  mutual   inductance of the loop.  The magnetic field of the long wire  depends  only  on the distance from the wire.  To find the magnetic flux  through  the  loop, we choose a strip a distance  x  from the wire with width  d x :   The mutual inductance is M  =  Φ B / I  =         ( μ 0 w /2p)  ln( ¬ 2 / ¬ 1 ). 6. We find the induced  emf from  å  = –  L  ? I /? t  = – (180 mH)(38.0 mA – 20.0 mA)/(340 ms) =       – 9.53 mV . The emf is opposite to the direction of the current, to oppose the increase in the current. 7. We estimate the inductance by using the inductance of a solenoid: L  =  μ 0 N 2 A / ¬  = (4p × 10 –7  T     m/A)(20,000 turns) 2 p(1.85 × 10 –2  m) 2 /(0.45 m) =       1.2 H . Use Word 6.0c or later to view Macintosh picture.
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Chapter  30,  p. 2 8. Because the current in increasing, the emf is negative.  We find the self-inductance from  å  = –  L  ? I /? t ; – 8.50 V = –  L [23.0 mA – (– 22.0 mA)]/(21.0 ms), which gives  L  =        3.97 H .
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