Page 348 - Schaum's Outline of Theory and Problems of Applied Physics
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CHAP. 27]                               MAGNETISM                                     333



        27.20. An air-core solenoid wound with 20 turns/cm carries a current of 0.1 A. (a) Find H and B inside the solenoid.
               (b) An iron core whose permeability is 6 × 10 −3  T·m/A is inserted in the solenoid. Find H and B now.

        27.21. A sample of carbon steel has a permeability of 0.01 T·m/A when the magnetic intensity is 75 A/m. (a) Find the
               magnetic field in the sample at this value of H. (b) Find the field at this value of H.





                               Answers to Multiple-Choice Questions


          27.1. (a)           27.8. (a)

          27.2. (c)           27.9. (b)

          27.3. (d)          27.10. (d)
          27.4. (c)          27.11. (b)

          27.5. (b)          27.12. (a)

          27.6. (a), (b), (c), (d)  27.13. (c)

          27.7. (d)          27.14. (b)






                               Answers to Supplementary Problems


         27.1. All three fields are detected in both cases.

         27.2. At first the mutual electric repulsion of the electrons causes the beam diameter to increase, but as they go faster the
               magnetic attraction becomes more significant and the beam diameter decreases.

         27.3. West; east

         27.4. The particle’s energy is not changed since the magnetic force on it is perpendicular to its direction of motion, and
               so no work is done on it by the field. The particle’s direction changes, however, and hence its momentum, which is
               a vector quantity, also changes.
         27.5. The closer the lines of force are drawn in a particular region, the stronger the field there.

         27.6. To the north

         27.7. (a) When the wire is parallel to B  (b) When it is perpendicular to B

         27.8. Such a loop does not tend to rotate when its plane is perpendicular to the direction of the magnetic field.
         27.9. 25 A

         27.10. 67 cm

         27.11. Since the fields are in opposite directions, B = 2 × 10 −5  T − 0.67 × 10 −5  T = 1.33 × 10 −5  T.
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