ANSWERS

QUIZ  8, CHAPTER 27 PROBLEMS : 1, 4, 6, 9, 15, 16, 17, 22, 23, 29, 35, 36, 37, 68
Turned in already.  

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

#4 and #9, concerning the force of a magnetic field on a wire current,  were reviewed in class. #12  was compared with #9 and found to be easier. That's because the  angle is constant during the integration along the loop---see  figure 27-41.
29.  was reviewed in class:  Use m*v2/r = |q|*v*B, where B , v and |q|  are magnitudes of the magnet field  velocity and charge. You want to solve for m*v in terms of symbols other than r. Here's how:  Let theta be the angle between the vertical and the radial line from the  center of the circle to the particle after it has risen  distance d and traveled horizontally L shown in figure. Clearly ,
r*sin theta = L and also  d = r - r*cos theta. Note: sin2theta + cos2 theta = 1. You can also do the problem using the Pythagorean Theorem to find r in terms of symbols. This can be seen by substituting into the trigonometric identity. 

sin2theta + cos2 theta = 1.

Using  r*sin theta = L and   d = r - r*cos theta, we get:

(L/r)2 + (r - d/r)2 = 1.  Clear the fraction by multiplying both sides by r2. 

You should be able to simplify and eliminate r in order to produce the result for m*v in terms of d, L, |q| and  other possible existing relevant constants..
6. mg = ILBsin 90 = ILB
Now the gravitational force, or weight,  points down. Thus the magnetic force points up. The direction of the current  I depends on the  magnetic field direction. For example suppose  the horizontal wire   was placed parallel to the page and in a horizontal uniform  field pointing  out of the page. In that case the current would flow leftward along the page. 

Find I , after re-writing the mass in terms of copper's mass density and the wire's volume. The volume is the product of  length L and  circular cross-sectional area computed from said diameter. 
15.  m*v2/r = |q|*v*B. Find B.
16. The following two practice problems a. and f. are first done  assuming a positive  charge. Reversing the direction of the force  in each case gives the  result  for a negative  charge. (a)  For practice, point your right fingers outward and bend them downwards toward vector B. Your right thumb points right ward along the page in the force direction. To get  result  for a negative  charge,  just reverse the direction of the force.     (f) Point your right fingers leftward along the page with your thumb upward. That is the only position allowing you to wrap you right finger into vector B. Your right  thumb points upward in the force direction. To get  result  for a negative  charge,  just reverse the direction of the force . 

You should be able to do the other parts on your own.  
17.  See the previous problem.
22. Use the cross product rule for the mutually perpendicular  unit vectors i , j and k along the x , y and z axes. Here's a suggestion.  Do the problem as if the charge was a proton, which is positive. Then  reverse the direction to get the  force on the the electron, which is negative. 
Use  these rules: 

i x i = 0, j x j = 0  and k x k = 0,  i x j = k, k x i = j  and j x k = i . Use these rules in combination with the distributive property of multiplication. Note: i x j = - j x i, etc.  Also see cross product examples in section 11.2.  Obviously in this  problem the force will  be along the z axis we only have products of i and j. The question is, is it up or down the z axis? Here we go:

(7.0 i -  6.0 j) x (-0.80i + 0.60 j)  = (7.0)(-0.80) i x i +(7.0)(0.60) i x j  + (-6.0)(-0.80) j x i  + (-6.0)(0.60) j x j. 

Simplify. 
23. Use  m*v2/r = |q|*v*B. Problem is we do not know v. But we can use the definition of kinetic  energy in order to find it. Here we go: KE = (6.0 MeV) x (106 eV/MeV) x (1.6x10 -19 J/eV) . After this conversion, set KE = (1/2)m*v2 to find speed v. With that,  find r  using the first equation. 
35. The work required by  you to rotate the dipole   equals the gain in potential energy. Below, the symbol u represents the magnitude of the magnetic moment. 
Work by you = gain in potential energy = - -u*B*cos thetaf   - (- u*B*cos thetai ) = -u*B*cos thetaf   + u*B*cos thetai , where thetaf  and thetai are the final and initial angles between the dipole moment and  magnetic field vector.
Just substitute specific given values for thetaf  and thetai  into the above formula .  

Note:  Work by field = -(gain in potential energy) . 
Thus,  Work by you = -(Work by field) . 
37. The Earth's magnetic field points 66.0 degrees below the horizontal. Looking downward, a  current flows through the loop that lies flat on the ground.  Let into the page be downward toward the ground. Imagine a clockwise current in a circle lying flat on the page .  Wrap your right fingers  clockwise and your thumb points into the page, the same direction as the magnetic dipole moment vector.   Thus the angle  between the magnetic dipole moment and the Earth's magnetic field is 90 - 66 = 24 degrees. Use formula 27-9 to get the torque magnitude. The magnetic dipole moment initially points downward then twists upward toward the Earth's field.  The edge of the loop furthest from the  Earth's geographic North sinks downward.  Thus the South edge of the loop sinks down. Which side rises upward, the North , South,  East or  West edge of the loop? 
68. This is a nice practical problem. (See also  69, which will be discussed in class.)  The loop will rotate in the manner shown since the magnetic dipole moment vector wants to align itself along the B-vector. Now, in fig. 27-52, the force  on the upper horizontal wire segment  points up,  and the force  on the bottom horizontal wire segment points down. These two forces reach maximum magnitude when the plane of the loop is perpendicular to the magnetic field vector. In the picture, the loop is approaching the position where its plane is perpendicular to vector- B. At that location, the force on the upper and lower horizontal wire segments  each have magnitude F =  I*L*Bsin 90 = I*L*B. These two forces provide a tension force on the vertical segments  as if you were pulling them at each end. According to section 12- 4 and section 12-5, there is a maximum tension at which the wire fractures i.e. breaks in two.  F/A = 2.0x10 6 N/m2  = maximum (force/area) before wire  breaks.   Taking into account the safety factor of 10, F = 10*I*L*B . Thus 10*I*L*B/area = 2.0x10 6 N/m2 , where area = pi*d2/4 and pi = 3.14...Find d.
(b) Find R = (resistivity)*L/(area).