Download Chapter 9 – solution

Student’s Name:……………………………
Student’s No:………………….
Electromagnetic II Discussion
Electrical Engineering Department
Faculty of Engineering
The Islamic University of Gaza
Chapter 9 :Maxwell’s Equations
Deadline : The next discussion
Q1) Choose the correct answer:
1-When there is a change in the magnetic field in a closed loop of wire:
a. a voltage is induced in the wire.
c. electromagnetic induction occurs.
b. current is made to flow in the loop of wire.
d. all of these
2-The concept of displacement current was a major contribution attributed to:
a. Faraday
b. Lenz
c. Maxwell
d. Lorentz
3- A loop is rotating about the y-axis in a magnetic field B = B0 sin (wt) ax Wb/m2 , The
voltage induced in the loop is due to :
a. Motional emf
b. Transformer emf c. A combination of motional and transformer emf
4- The flux through each turn of a 100-turn coil is (t3 – 2t) mWb, where t is in seconds. The
induced emf at t = 2 s is:
a .1 V
b. -1 V
c .4 mV
d.0.4 V
5- Which of the following statement is not true of a phasor:
a. It may be a scalar or vector
b. It is a time – dependent quality c. It is a complex quality
6- The velocity of wave propagation depends on:
a. Permittivity of material b. Permeability of material
c. a and b
Q2) True or False:
1. ( T ) For a field to be qualified as an electromagnetic field, it must satisfy all four Maxwell's
equations.
2. ( F ) The magnetostatic fields are usually produced by static charges.
3. ( F ) The electrostatic field is usually produced by the motion of electric charges with
uniform velocity.
4. (F ) According to Faraday’s experiment, a steady current produces a magnetic field.
5. (F) A time-varying magnetic field produces an induced voltage called (electromagnetic
force)
6. (F) The Maxwell’s equations for time varying fields show that the time-varying E field is
conservative.
7. (F) At high frequencies, the displacement current density is usually neglected compared
with conduction current.
8. ( T ) Without the displacement current electromagnetic wave propagation would be
impossible.
9. ( T ) The concepts of linearity, isotropic and homogeneity of a material medium still apply
for time – varying fields.
Q3) a conducting rod moves with a constant velocity of 3az m/s parallel to a long straight wire
carrying current 15 A as in figure. Calculate the emf induced in the rod.
Hint: B=
o I
a due to long conductor
2
Solution :
Vemf = -9.9 uV
Q4) Consider the loop of Figure. If B = 0.5az Wb/m2, R = 20 Ω, ℓ = 10 cm, and the rod is
moving with a constant velocity of 8 ax m/s, find
a)The induced emf in the rod
b)The current through the resistor
c)The motional force on the rod
d)The power dissipated by the resistor.
Solution:
a)Vemf = -0.4 V
b) I = -20 mA
c) F = -.001 ax N
d) P = 8 mW
Q5) A
triangular wire loop has its vertices at the points (0,0,0), (2,0,0) and (0,0,4) with
dimensions in meters. A time-varying magnetic field is given by 𝐁 = 4𝑡 ayWb/m2. If the wire
has a total distributed resistance of 2 Ω, calculate the induced current and indicate its
direction.
Solution:
𝑉𝑒𝑚𝑓 = −
𝑑𝐵
𝑑𝑠
𝑑𝑡
𝑑𝐵
= 4 𝑎𝑦
𝑑𝑡
𝑑𝑠 = 𝑑𝑥 𝑑𝑦 𝑎𝑦
∴ 𝑉𝑒𝑚𝑓 = −16 𝑉
𝐼 = −8 𝐴
Q6) The
conducting wire shown in the figure rotates at 4 revolutions/second. Assume the
wire is located along the x-axis at time t=0. Calculate the induced voltage Vemf.
Solution :
𝑉𝑒𝑚𝑓 =
𝑢=
𝑢 × 𝐵 . 𝑑𝑙
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 4 2𝜋𝜌
=
= 8𝜋𝜌 𝑎∅
𝑡𝑖𝑚𝑒
𝑠
𝐵 = .01 𝑎𝑧
𝑢 × 𝐵 = .08𝜋𝜌 𝑎𝑝
𝑑𝑙 = 𝑑𝜌 𝑎𝜌 , .02 < 𝜌 < .08
𝑉𝑒𝑚𝑓 = .754 𝑚𝑉
Q7) In free space
H = ρ(sin φ aρ +2 cos φ aφ) cos 4 x 106 t A/m
Find Jd and E using Maxwell equations
Solution :
𝛻 × 𝐻 = 𝐽 + 𝐽𝑑 ,
𝐽 = 𝜍𝐸 ,
𝐽𝑑
𝜍 = 0 ∴ 𝛻 × 𝐻 = 𝐽𝑑
𝑑𝐷
=
,𝐷 = 𝜀 𝐸
𝑑𝑡
Q8)
A wire is connected to the inner shell (and insulated from the outer so it doesn't short
out) and fed with a 1 GHz ,1 Amp sinusoidal current.
a) Explain how the current can pass through the air gap between the inner and the outer
shells.(use Maxwell equations)
b) What is this current called?
c) Is it caused by moving electrons?
d) How large is this current?
e) Would a DC current also pass through this air gap?
Solution:
a) 𝛻 × 𝐻 = 𝐽 + 𝐽𝑑
b) Displacement current
c) No
d) 𝐽𝑑 =
e)No
𝑑𝐷
𝑑𝑡