Chapter 9 CQ

• Consider a solenoid oriented on a horizontal surface from north to south. We know that self-induced emf  in this solenoid is not zero and directed from south end of solenoid to north end. We can conclude from this fact that current though this solenoid is:

  not enough information
• Consider a solenoid oriented on a horizontal surface from north to south. We know that big constant in magnitude current is passing through this solenoid from the south end to north end. We can conclude that self-induced emf in this solenoid is:

  Is zero
• You have two solenoids of the same length, same density of turns, but solenoid #2 has twice the diameter of solenoid #1. We can conclude that inductance of solenoid #1 is

  quarter of inductance of solenoid #2
• You have two solenoids of the same lengths, diameter and density of turns. But current through solenoid #1 is twice as big as current through solenoid #2. We can conclude that inductance of solenoid #1 is:

  Same inductance as solenoid #2
• A long solenoid has the cross-sectional area A. Coil #1 with the cross-sectional area 2A is coiled around solenoid. Coil #2 (with the same # of turns as coil #1) with the cross-sectional area A/2 is placed inside of solenoid, so that turns of the  solenoid and turns of the coil are parallel to each other.  Coil #1 has mutual inductance with solenoid  which is

  2 times mutual inductance of coil #2 with solenoid