E. Electromagnetism Homework
20.1 Induced Emf’s: Faraday’s Law and Lenz’s Law
1-3, 8, 15, 17, 22
8: Remember q = angle between the NORMAL to the plane of the loop and the B field.
15: Remember that DF = Ffinal - Finitial where Ffinal = 0 so DF will be a negative value.
17. Recall that Î = IR (first find Î)
22: Look at conceptual example 20.3 on p. 640. Velocity needs to be in m/sec
20.2 Generators and back Emf:
28 – 33, 41, 43
31: No free lunch!
32: Draw a Cos graph of this. At t = 0 sec you have max Î. Therefore, at T/4 you have zero and at T/2 you have -Îmax. Also, since 60 Hz is the frequency (f), the period (T) = 1/60.
33: Î=NBAw where w = 2pf
41: Îback = V-IR
43: a) Îback = max = V-IR b) Is there a back Emf on start up? C) Recall that resistors in series are additive and you already have 1.5W resistor in the motor. So, how much more W will you need to add to this 1.5W?
20.3 Transformers and Power Transmission
44 – 47, 51
51: Np/Ns = Vp/Vs
Also, Read this passage and answer the following question:
You have all seen those little black adapters which plug into your calculators or small appliances. These are STEP DOWN transformers. They do this by not only having a step down transformer inside of them, but also a series of resistors. Recall from the chapter that in a STEP DOWN transformer, the voltage is lowered, but the current is RAISED. Therefore, you want to make your transformer have some resistive nature in order not to burn out your device with too much current.
I have a STEP DOWN transformer for Megan’s re-chargeable Hot Wheels Barbie Jeep. It says: INPUT: 120V (AC) at 60 Hz 12W and OUTPUT: 6.0V(DC) 800mA
a. What is the current that it draws from the electric company when I plug it in? (P = IV)
answer: 12W/120V = 0.1A
b. What is the current that it sends to the Jeep? (the 800mA value is not correct as we will see in a moment). Ip/Is = Vs/Vp
answer: 0.1A/Is = 6V/120V so Is = 2 Amps
c. When you read the OUTPUT current, you find that it is 800mA (0.8 A). but this is much less than the amount you got in part b. If you assume the current you got in part b is flowing at a voltage of 6V, what is the “theoretical” power produced in the OUTPUT? What is interesting about your answer? (P = IV)
answer: P = 2 amps x 6 volts = 12 Watts. That is the same as the input power.
d. We know by reading the transformer that it produced 800 mA at 6V. What power is ACTUALLY produced? (P = IV)
answer: P = 0.8 A x 6 V = 4.8 W
e. What power was lost? (Plost = Ptheory – Pactual)
answer: Plost = 12 W – 4.8 W = 7.2 W
f. What is the efficiency of this transformer? Obviously it isn’t 100% efficient. (Efficiency = Pout of the secondary loop/Pout of the primary loop) – when power in = power out, you have 100% efficiency.
Answer: E = (4.8 W/12 W) x 100 = 40%
g. What resistance inside of the transformer would account for this power loss?
(Plost = I2primary Rprimary)
answer: 7.2W = (0.1A)2R
h. Where does the lost wattage of power go?
Answer: lost as heat
j. If the Power lost = 0, then what would the internal resistance of the transformer be?
Answer: zero! No resistance at all because all of the 0.1A of current in the primary coil would get through.
k. If there are 100 coils of wire around the primary coil, how many are around the secondary? (according to the voltages listed on the transformer box)
answer: Vs/Vp = Ns/Np = 6V/120V = 0.05 : 1
100 x 0.05 = 5 coils.
8: a) 0 b) 2.7 x 10-3Tm2 c) 4.5 x 10-3Tm2
30: a) plane of the loop is ^ to the B field because the magnetic field flux is maximum at this angle and any changes in rotation to the loop will cause only a small change in the magnetic field. b) plane of the loop is parallel to the B field because the magnetic field flux is minimum and any change in the loop’s orientation would cause a great change in the amount of magnetic field going through the loop.
32: a) 1/120 sec b) 1/240 sec c) 1/60 sec