Quiz  20 ( ANSWERS)   CQ 9, 10 Problems  3, 4, 5, 6, 11, 12, 13, 14, 15, 16, 19, 27, 28, 29, 39, 33, 40, 41, 42, 43, 44, 45, 46
TURN IN: CQ 9, 10;  Problems 4, 6, 12, 14, 16, 28, 40, 42,  44, 46

Problem DISCUSSIONS below:
error log #1: See correction to 14: 1 MeV = 1.6x10-13 Joules
error log #2: See correction to 42: d' is the symbol for the distance between wires;  d = wire diameter.
4. See section 20-3, equation 20-1. The length, angle, B and I are given. 
6.   I*L*Bsin(theta) =  (0.35)*I*L*B . Find theta.

12. 
(a) The vector-B is perpendicular to vector-v since the force magnitude is the maximum possible (sin 90 = 1) according to  the formula  for the magnetic  force on a  charge moving with speed v.  Re-read section 20-4.  The vector-B also lies in the plane of the page since the the outward magnetic force is perpendicular to the plane containing vector-v and vector-B. Thus vector B is either upward  or downward . Choose the direction according to RHR-3 in Table 20-1 for the  force  on electric charge  + q . Use  similar reasoning for parts (b) and (c). 
14. Find the speed v from the formula for kinetic energy: (1/2)*m*v2  = 5.0 MeV. Note you must convert MeV into Joules:
1 MeV = 1.6x10-13 Joules. Once you find speed v , then read about circular motion  in a uniform magnetic field. See figure 20-16 and example 20-5 for an electron.  Note a positive particle would feel an force in the opposite direction and would thus move counter-clockwise.
16. The electric force is equal and  opposite to the magnetic force since the net force is zero on the un-deflected electrons. You can get help visualizing the problem from  # 12. Let  figure 20-52 (a) represent  the motion of an electron. The electric force on the electron  would have to point into the page. 


Since the electron is negative, two observations should be made. (1) The magnetic field must point upward (instead of upward) along the page if the magnetic force points out of the page. (2) The  electric force is opposite  the electric field. Thus the electric field points out of the page.

If the net force is zero, e*v*B = e*E, where e = electron  charge magnitude. Solve for v.

If the electric field is turned off, the force will be purely magnetic. See problem 14 hint  to find r for the consequent circular motion.
28. In all three cases, draw a straight line segment from the wire to  points C, D or E. The magnetic field vector will be perpendicular  to that line segment in each case.   See RHR-1, Table 20-1. 

Here's a hint to help you through all three parts. At  point C, according to RHR-1, your thumb points out and your right fingers form a  counter -clockwise circle. Thus at point C, the magnetic field points horizontally  left.
40. This is a cool problem. The following equations are true for the final magnetic field due to the earth and the wire. In class, they will be discussed and diagrammed in detail.

(1) Bfx  = Besin23
(2) Bfy  = Becos23 - u0I/(6.28r ).

Here Be = magnitude of Earth magnetic field.
u0I/(6.28r ) is the magnitude of the field due to the wire. 6.28 means 2*pi = 2*(3.14159...). Use the pi button on your calculator.  r = 0.12 m.

We want to solve for I.  We have three unknowns Bfx, Bfy   and I but only two equations.  But an additional equation is

(3) Bfx  = Bfy*tan55.

Solve (1), (2) and (3) for I.

42.

(a) Read section 20-6. See figure 20-25 and example 20-11.  The gravitational force on the bottom wire is down, so the magnetic force must be up. You want to solve for current-I2, both magnitude and direction. The magnitude  equation will be :
u0I1*I2*L/(6.28d') = mg, where L is the common length of the wires and I1 = 48 (A), and d' is the distance between wires.   Note that m = (copper density)*pi*L*d2/4, where  d = diameter.  Copper density (convert to  kg/m3)  can be found in the book or on the internet.  After making this substitution, L will cancel out. Find magnitude I2.  Find the direction with the help of figure 20-25 and example 20-11. 
(b) If the bottom wire was displaced slightly lower, the gravitational force would be larger than the magnetic force and the wire would continue downward.  Is this stable or unstable equilibrium? Look up the definitions in the book or on the internet.
(c) Re-do the problem  with the copper wire above the first wire.
44.  Assume the two parallel currents flow out of the page. You can have the wires horizontally displaced from one  another.  Let wire B be positioned to the right of wire A,  i.e. wire A is on the left and wire B is on the right.  When you hand in this problem, please show a diagram reflecting this set up. 

(a) Wire A is on the left and wire B is on the right.    To find the direction of the field at A due to B, your thumb points out in the direction of current B. Your fingers will wrap in a counter-clockwise direction. Thus , at wire A, the field from B points down. 

(b) Following an analogous procedure as in part (a), you will find that at wire B, the field from A points up. 

(c) The formula for the magnitude B  of the magnetic field due to a long wire is B =    u0I/(6.28r ),  where r is the distance from the wire. See problems 40 and 42. In this case r = 0.15 m.  Since the two currents are not equal,  what can you say about each field magnitude discussed above? Are they equal? 

(d) To get the force on each wire, see section 20-6. See example 20-10. Remember that parallel current attract and opposite currents repel.  Note that since the wire lengths are not given, you will have to find the magnitude of the force per unit length.  See the bottom of page 566 for a discussion. Just divide the force by the common assumed length L = LA = LB .  So for instance in example 20-10 you would divide by L = 2.0 m to get the force per unit  length.  
46. This problem was discussed in class and will be graphically explained during this week's review sessions. 
error log #1: See correction to 14: 1 MeV = 1.6x10-13 Joules
error log #2: See correction to 42: d' is the symbol for the distance between wires;  d = wire diameter.
More hints later!