QUIZ 8~   CH. 28
EXERCISES/PROBLEMS
Magnetic fields --a basic description  of field lines--they form loops with no beginning or end with an emphasis on  magnetic forces:  Magnetic forces only act on moving charges. THE STARTING POINT OF THE ENTIRE DISCUSSION FOR CHAPTERS 27 AND 28  IS THE BEGINNING OF SECTION 27.2: Magnetic fields are caused by moving atomic  charges (i.e.,  electrons)  and only moving charges experience the magnetic force.  IN THIS CHAPTER WE EMPHASIZE  MOVING CHARGES OR CURRENTS AS SOURCES  THE MAGNETIC FIELDS. 
* indicates the problem is turned in.  EXERCISES: Field of a moving charge:
 3,  4*,  8*, Field from a current element or segment or long wire: 30*, 66*(try 69),  76, 12*, 19, 24* (try 56), force between parallel wires: 28* (try 63, 64*), 29*, Magnetic field of coil: 31, 32*, 68* (a) only,  Ampere's Law: 36, 37*, 38*, 39*, solenoid 41*    
 DISCUSSIONS OF PROBLEMS NOT DONE IN CLASS FOLLOW 
 Done in class: # 4 and #39
8. See # 4,  The proton produces a field at the origin that is INto the page. The electron produces a field that is also INto the page.
30. EXTREMELY IMPORTANT ~  can be done using equation 28.6.
First off for that equation in differential form, we know the field at  center point P is IN to the page.  Field magnitude Bp = B_arc + B_line  = B_arc since B_line = zero from equation 28.6. That's because the dl vector and the r "hat" unit vector are in the same direction making the cross product = 0. Now. differentially,  dB_arc =  
uo*I*R*dtheta/(4*pi*R2) = uo*I*dtheta/(4*pi*R)  where dl = R*dtheta.  Integrate with respect to  theta between zero and pi. to get magnitude; the vector itself points INto the page.
66. See the previous problem. Magnitude Bp = B_arc_lower - B_arc_upper, where B_arc_upper is the field due to the upper arc and "arc_lower"  refers to the lower arc. 
The field  from the upper arc points IN. The field from the lower arc points OUT. The the magnitude of the filed from lower arc is larger is since the distance is closer. Hence the net field is OUT. 

Compute the integral of the following by integrating with respect to theta from 0 to pi. :    uo*I*dtheta/(4*pi*a)   - uo*I*dtheta/(4*pi*b).
12. This is practice with equation 28.6 or 28. 7  applied to current elements; see example 28.2.

 At point P,  the total field is IN. Bp is IN, since the field from each current element is IN. Thus,
Bp = uo*(I1 + I2*)(0.0015 m)*sin theta/(4*pi*r2), where r = 0.08 m.

24. The two right ward currents  create a NET   field that is  OUT : The bottom wire produces a field that in OUT and the top wire creates  a field that is IN but with smaller magnitude--thus net field from the sum is OUT.  The upward 10.0 A current creates a field that is IN. You should be able to see that the NET FIELD from all of these three currents is OUT.  Thus the field from the unknown current I must be IN, and that current flows downward along page. Note: Use equation 28.9 or the same equation on page 970 (Ampere's Law) for time , each time using the distance (a/1.41), where 1.41 means the square root of 2 and side a = 0.40 m.