|QUIZ 21 CHS. 21, DISCUSSIONS POSTED|
SELECTED DISCUSSIONS TO exercises/problems
YOU ARE RESPONSIBLE FOR ALL PROBLEMS EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.
|CH. 21 MULTIPLE CHOICE TBA - Exercise/problems FLUX
1; FARADAY'S LAW 4, 5, 7; LENZ'S LAW 13, 15, 20; MOTIONAL EMF 21,
25; SELF INDUCTANCE 33; TRANSFORMERS 38; FIELD ENERGY
R-L CIRCUITS 48; LC CIRCUIT (AIN'T NUTHIN' BUT A SPRING ATTACHED TO A MASS W/O NO FRICTION) 54, 55
|TURN IN: 1, 5, 13, 15, 21, 25, 33, 38, 44, 48, 54.|
|* DISCUSSIONS PROVIDED.|
|1. (a) flux = BA*cos 0 (b) flux = BA*cos53.1(c) flux = BAcos 90 = 0.|
|5. Let | E | represent the
magnitude of the induced voltage aka emf. |E| = N*|change in
flux/change in time| . Also , I = |E|/R. Now ,
|change in flux| = |change in B|*area. I = 0.150 (A) and area =
pi*r2. where r is given. N = 200. Find
| change in B |.
|13. See lab # 3. and figure 21.15. (a) If the external B-vector points in and increases in magnitude, the induced magnetic field vector-Bind will point out and thus the current is counter clockwise according to figure 20.38. Remember you can find the induced vector-Bind by wrapping your right fingers in the direction of I in the loop and your right thumb points in the direction of induced vector-Bind along loop axis.|
|15. The solenoid is like a magnet and the North pole is at its upper end in the figure shown. In the figure, the external B field from the solenoid points up, out from the North pole and is increasing in the loop as someone moves the electro-magnet upward. Looking upward, the external field points in as in problem 13 (a) and is also increasing in magnitude; from that info, is the current clockwise or counter-clockwise?|
|21. Think of the bar as lying parallel to Highway 580, which has an EAST-WEST orientation. The magnetic field points upward toward the sky. Now, move the bar horizontally northward toward Oregon. See figure 21.18 (a) and pretend rightward, the direction of motion, is Northward and in the page is up toward the sky. In this interpretation, the top of the bar is East and the bottom is West. We see the force on the charges would be up (East) in the figure.|
|25. (a) Let | E | represent the magnitude of the induced voltage aka emf. Thus, |E | = BLv. (b) See figure 21.18 (b) to reach your conclusion. (c) See example 21.5. Since a current flows, there is a force of magnitude ILB as shown in section 20.5. That force points left. Note: I = BLv/R.|
|33. See class notes distributed in class. See figure 21.25.|
|38. See Group ICQ, and equations 21.15 and 21.17.|
|44. (a) Energy = (1/2)*LI2. Find L given Energy. (b) See the result of #36: (i) L = uo*A*N*(N/L), where N/L = 10/mm. Thus, (ii) L = N/(10/mm). Substitute this latter expression (ii) into the equation for L (i) and find N. Then go back to the latter equation (ii) to find L.|
|48. See equation 21.25: (a) time
constant = L/R. (b) Imax = 12.0 (V)/R. (c) Plug in the time t
into equation 21.25.
(d) Energy max = (1/2)*LImax2 .
|54. Please also try #55; you may be tested on it. Meanwhile here is 54: Use equation 21.30.|
|55. (a) Total energy = initial energy =
(1/2)Q2/C, where Q = Qmax is the initial charge =
(b) From conservation of energy, (1/2)*LImax2 = 1/2)Q2/C, where Q is the initial charge = C*(initial voltage) . Note when Q = 0, I =Imax and when I = 0, Q = Qmax .