ANSWERS

QUIZ 9   , CHAPTER  28 PROBLEMS : 5, 18, 21, 22, 26, 27, 34, 35, 37, 38, 39.

Turn in only: 18, 26, 34; Discussions  provided soon, but try and read examples for help.  

Simulation: Click here for the motion of an electron in a uniform electric field.

 18.There  is a naive model and a rigorous model for this problem.

Let us try the naive model to avoid computational difficulties. First off check out the situation and look at the picture conceptually. Forget  the math for a second and look at the picture . We see three pairs of interactions: 1. attraction between the long wire and the side of the loop closest to it (top side) , 2. repulsion between the long wire and the side of the loop farthest from it ( bottom side in the figure) and finally, 3. the repulsion  between the two sides of the loop; it turns out that when you compute the NET force on the loop, these latter contributions cancel out. Let's gets to work:

Force on top side of loop is ILBsin90, where clearly the angle is 90 degrees between the magnetic field from the long wire and the direction of the current in top side. Now, the field magnitude (from Chapter 28 and 27)  B = uoI/(2*pi*r), where pi = 3.14, r is the perpendicular distance from the wire and uo is 4*pi *10-7 T·m/A. Thus:

Force (top) = I*L*uoI/(2*pi*r), up, ignoring  the interaction with the bottom loop.  
Force (bottom) = I*L*uoI/(2*pi*r'), down, ignoring  the interaction with the top  loop.  Note that r' is the distance between the long wire and the bottom side. 

The net force has a magnitude equal to the positive difference between the previous two force  values.  
 26. B = uo*n*I, where uo  =  4*pi *10-7 T·m/A, and n is the number of turns per unit length and I is  the current. Thus n = N/L where N is the number of turns and L = 0.32 (m) . Note the diameter is irrelevant . Solve for N. 
34. See example 28- 13. That example is critical ! I also actually did this problem in class; it uses  the Biot Savart law and certain geometric simplifications. There are 4 sections to the loop---two straight sides and two curved, and outer and an inner. The straight sides do not contribute to the field at the loop center because of the cross  product in equation 28-5...Finishing this we get: 

For  the field from the top curved side, the direction is IN, following example 28-13. The magnitude is given by:

  uo*I/(4*pi*R2) *( R2*θ  ) .

The contribution from the bottom curve has a similar form, except the radius will be different, and the direction will be OUT.....But the angle will be the same ! Take the difference to compute the net magnitude, subtracting the smaller magnitude from the larger one.  
39. WE HIT THIS  PROBLEM FROM MANY ANGLES, AND ONE WE SUGGESTED SEEMS TO
WORK, LOGICALLY AND NUMERICALLY,  IF YOU FOLLOW THE RULES SET DOWN BY
THE PREMISE.

Here is the corrected approach up to the point where you perform the
integral that gives you the answer  in  back of the book:
Basically you divide the disk into differential rings of current,
each having its own magnetic dipole moment. Then integrate to get the
total moment. In other words,  integrate differential du to get u,  the total dipole moment.

Divide the disk into differential rings, of thickness dr and
circumference 2*pi*r. First off, you need to get the differential
dipole moment du = dI*area, where dI is the differential current in one such ring
and area = pi*r2  = circular area bounded by a ring. Note: dq =
[Q/(pi*R2) ]*2*pi*r*dr = (charge density)*dA, where dA =
differential area of a single ring = 2*pi*r*dr .

Also note: dI = dq/period, where period = 2*pi/w  and w is the angular speed.

Now, du = [Q/(pi*R2) ]*[ 2*pi*r*dr ] * [ w/(2*pi) ] * [ pi*r2 ] .
Clearly we see  pi cancels out,  suggesting  the final
answer in  back of the book.  WHEN YOU SIMPLIFY ALL THE FACTORS, you get for the differential moment:

du = [ Q*r3*dr/R2 ] * w.  Clearly when you integrate this from 0 to
R you get the answer in back of the book, including the elusive
factor of 4 !

MORE DISCUSSIONS OF PART (B) AND (C) PROVIDED  BELOW , but also see pp.
744-5 and the discussion emanating from example 28-12. 

(b) You would have to get the result for dB from one differential  ring as shown in equation 28-7a  for x not necessarily  much larger  than R.  This dB is caused by the differential dipole du that would be in the numerator  of equation 28-7a. Then you would integrate the dB's from  all the  rings  from 0 to R.  Then take the limit as x >>R for part (c) . Stay tuned.

dB = [ Q*r3*dr/R2 ] * w /(r2 + x2)3/2.

This integral can easily be  performed by parts. Look at the basic integrand form here: r3*dr/(r2 + x2)3/2, neglecting the constants which you can add back in at the end of the computations.  Let  INTEGRAL be used for the integral sign and the limits of integration between 0 and R .

INTEGRAL  r3*dr/(r2 + x2)3/2     

 =   INTEGRAL  r2*r*dr/(r2 + x2)3/2  =   - R2* (R2 + x2)-1/2    +  INTEGRAL  (r2 + x2)-1/2 *2r*dr, where on the right hand side the limits of evaluation and integration are 0 and R.
( THE  SECOND TERM ON THE RIGHT, THE INTEGRAL, IS  EASILY EVALUATED SINCE THE INTEGRAND IS A PERFECT DIFFERENTIAL.  PLEASE  VERIFY THESE FORMULAS AFTER CONSULTING YOUR CALCULUS BOOK AND USING YOUR EXPERTISE TO REPRODUCE THEM;  do not just copy. )

(c) Finally evaluate you formula in the limit of  R << x < infinity.