Download ece221h1s: electric and magnetic fields

ECE221H1S: ELECTRIC AND MAGNETIC FIELDS
PROBLEM SET #12
TRANSFORMER, MOTIONAL, AND TOTAL EMFS
Readings (Week #12)
Ulaby Textbook: Chapter 6, pgs. 288 – 299 (Secs. 6-3: The Ideal Transformer, 6-4: Moving
Conductor in a Static Magnetic Field, 6-5: The Electromagnetic Generator, and
6-6: Moving Conductor in a Time-Varying Magnetic Field)
Example Problems: Faraday’s Law 1 – 3
Workbook (Knight):
Chapter 34:
1, 2, 3, 4, 5, 18
Textbook (Ulaby):
Note, in the problems below you can ignore the effects of any self-inductance of the
loop. This means that you only have to consider the applied magnetic field, so you can
ignore the induced magnetic field if there is one (i.e., if the loop is closed such that an
induced current can flow). This applies particularly to the questions with closed loops,
6.7, 6.11, 6.12, and 6.13)
Chapter 6 Problems
Exercise 6-3: Solution on CD
6.9:
Exercise 6-4: Solution on CD
6.10:
6.7:
6.11:
6.8: (a)
(b)
=
= 5.5 sin(377 ) [V]
6.12:
= 22.6 [μV]
=
−
= −236 [μV]
= 0.1 [A]
6.13:
Supplementary Problems:
1. Try the posted Faraday’s Law Example #1 (ECE221Ex_FaradaysLaw1.pdf). The full solutions
are given.
2. Q2b from the 2009 Final Exam:
(8 marks)
(2 marks)
(b) A single-turn circular loop of wire, which lies in the xy plane, has a radius of a = 5 cm
and has a small gap cut into it. Attached to it is a handle, which allows the loop to be
rotated about the x-axis. The loop lies in the presence of the earth’s magnetic field,
which is approximately given to be B = 0.5 x 10-4ây T.
(i) Determine the number of rotations per second, such that the peak value of the
induced emf is 120 Volts.
(ii) Suggest two changes that you could make to the above system which would reduce
the required number of rotations per second.
1
Supplementary Problems (cont’d):
3. Q5a from the 2008 Final Exam:
A circuit which coverts mechanical energy into electrical energy is shown below. It consists
of a conducting sliding bar which oscillates over two parallel conducting rails in a
sinusoidally varying magnetic field =
cos( ) . The bar oscillates due to the two
identical springs that connect it to either side of the circuit. The position of the bar is
.
described by ( ) = 0.3 1 − 0.8cos( )
[m]. The circuit is connected to a light bulb
with a resistance of 0.1 Ω.
y
l = 60 cm
w = 20 cm
Light bulb with a
resistance R= 0.1 Ω
B
x(t)
x
(12 marks)
a) Given that ω = 120π [rad/s], determine the strength of the magnetic flux density, B0, which
is needed to produce a current through the light bulb of magnitude 1 mA at t = 50 ms. Indicate
on the diagram above, the direction which this 1 mA current flows through the light bulb at
this instant in time (t = 50 ms). You may assume that there is no magnetic flux through the
loop that is created by the leads which connect the light bulb to the rails. As well, the only
resistance in the circuit is that from the light bulb, and the circuit’s inductance is negligible.
2