EEE 241 QUIZ 5-8 COMBINED FUNDAMENTALS OF
ELECTROMAGNETICS ARIZONA STATE UNIVERSITY
LATEST UPDATED REAL FINAL PRACTICE EXAM WITH
ALL POSSIBLE COMPLETE DETAILED QUESTIONS
AND 100% CORRECT VERIFIED ANSWERS FULLY
SOLVED WITH CERTIFIED RATIONALES PLUS
RELIABLE ANSWER KEY (100% COMPLETE
SOLUTIONS) 2026-2027 UPDATED BEST VERSION
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1. At room temperature, which of the following
correctly compares the electrical properties of an
insulator, a semiconductor, and a conductor?
A) A conductor has many free electrons, a
semiconductor has a moderate number of free
electrons, and an insulator has very few.
B) All three have nearly the same conductivity.
C) A semiconductor has more free electrons than a
conductor but fewer than an insulator.
D) An insulator and a semiconductor both have no
,mobile charge carriers.
A
Conductors possess an abundance of free electrons
in their conduction bands. Semiconductors have a
moderate number of thermally generated electron-
hole pairs. Insulators feature a wide energy band
gap, resulting in negligible mobile carriers at room
temperature.
2. Two large, parallel conducting plates are
separated by a distance of d = 2 mm. Each plate has
an area of A = 0.02 m². The region between the
plates is filled with a material with dielectric
constant εᵣ = 3. Find the capacitance (in pF, one
decimal place).
265.5 pF
C = εᵣε₀A/d = (3) (8.854×10⁻¹²) (0.02)/ (0.002) =
265.5 pF.
3. A parallel-plate capacitor has plate area A = 0.02
m² and separation d = 1 mm. It is connected to a
constant voltage source of V = 76 V. If the upper
,plate is allowed to move slightly while the voltage is
maintained constant, what is the magnitude (in μN,
one decimal place) of the attractive force acting on
the upper plate due to the electric field? (The
material between the plates is air.)
51.1 μN
F = ε₀AV²/(2d²) = (8.854×10⁻¹²) (0.02) (76²)/
(2(0.001) ²) = 51.1 μN.
4. An electron (q = −1.602 × 10⁻¹⁹ C, m = 9.109 ×
10⁻³¹ kg) enters a uniform magnetic field B = 3 mT
with speed v = 23 × 10⁶ m/s directed perpendicular
to B. What is the radius of its circular path (in mm,
two decimal places)?
40.9 mm
r = mv/(qB) =
(9.109×10⁻³¹)(23×10⁶)/((1.602×10⁻¹⁹)(3×10⁻³)) =
40.9 mm.
5. The Biot–Savart Law for a steady current I is given
by: B = (μ₀ / 4π) ∫ (I dl × r̂) / r². What is the physical
meaning of this law?
, A) It describes the force between two charges.
B) It calculates the magnetic field produced by a
steady current.
C) It defines the electric potential of a charge
distribution.
D) It describes the induced electric field due to a
changing magnetic flux.
B
The Biot–Savart law gives the magnetic field
produced by a steady current distribution.
6. Two dielectric materials with permittivities ε₁ =
2ε₀ and ε₂ = 5ε₀ meet at a flat interface. The electric
field in medium 1 is E₁ = 3x̂ + 4ẑ V/m, and the
interface lies in the x–y plane (normal to ẑ). Find the
electric field in medium 2 just below the interface.
Assume there is no free surface charge at the
boundary.
E₂ = 3x̂ + 1.6ẑ V/m
Tangential E is continuous: E₂ₓ = 3. Normal D is
continuous: ε₁E₁ₙ = ε₂E₂ₙ → 2(4) = 5E₂ₙ → E₂ₙ = 1.6.
ELECTROMAGNETICS ARIZONA STATE UNIVERSITY
LATEST UPDATED REAL FINAL PRACTICE EXAM WITH
ALL POSSIBLE COMPLETE DETAILED QUESTIONS
AND 100% CORRECT VERIFIED ANSWERS FULLY
SOLVED WITH CERTIFIED RATIONALES PLUS
RELIABLE ANSWER KEY (100% COMPLETE
SOLUTIONS) 2026-2027 UPDATED BEST VERSION
RATED A+ GRADE INSTANT DOWNLOAD PDF
1. At room temperature, which of the following
correctly compares the electrical properties of an
insulator, a semiconductor, and a conductor?
A) A conductor has many free electrons, a
semiconductor has a moderate number of free
electrons, and an insulator has very few.
B) All three have nearly the same conductivity.
C) A semiconductor has more free electrons than a
conductor but fewer than an insulator.
D) An insulator and a semiconductor both have no
,mobile charge carriers.
A
Conductors possess an abundance of free electrons
in their conduction bands. Semiconductors have a
moderate number of thermally generated electron-
hole pairs. Insulators feature a wide energy band
gap, resulting in negligible mobile carriers at room
temperature.
2. Two large, parallel conducting plates are
separated by a distance of d = 2 mm. Each plate has
an area of A = 0.02 m². The region between the
plates is filled with a material with dielectric
constant εᵣ = 3. Find the capacitance (in pF, one
decimal place).
265.5 pF
C = εᵣε₀A/d = (3) (8.854×10⁻¹²) (0.02)/ (0.002) =
265.5 pF.
3. A parallel-plate capacitor has plate area A = 0.02
m² and separation d = 1 mm. It is connected to a
constant voltage source of V = 76 V. If the upper
,plate is allowed to move slightly while the voltage is
maintained constant, what is the magnitude (in μN,
one decimal place) of the attractive force acting on
the upper plate due to the electric field? (The
material between the plates is air.)
51.1 μN
F = ε₀AV²/(2d²) = (8.854×10⁻¹²) (0.02) (76²)/
(2(0.001) ²) = 51.1 μN.
4. An electron (q = −1.602 × 10⁻¹⁹ C, m = 9.109 ×
10⁻³¹ kg) enters a uniform magnetic field B = 3 mT
with speed v = 23 × 10⁶ m/s directed perpendicular
to B. What is the radius of its circular path (in mm,
two decimal places)?
40.9 mm
r = mv/(qB) =
(9.109×10⁻³¹)(23×10⁶)/((1.602×10⁻¹⁹)(3×10⁻³)) =
40.9 mm.
5. The Biot–Savart Law for a steady current I is given
by: B = (μ₀ / 4π) ∫ (I dl × r̂) / r². What is the physical
meaning of this law?
, A) It describes the force between two charges.
B) It calculates the magnetic field produced by a
steady current.
C) It defines the electric potential of a charge
distribution.
D) It describes the induced electric field due to a
changing magnetic flux.
B
The Biot–Savart law gives the magnetic field
produced by a steady current distribution.
6. Two dielectric materials with permittivities ε₁ =
2ε₀ and ε₂ = 5ε₀ meet at a flat interface. The electric
field in medium 1 is E₁ = 3x̂ + 4ẑ V/m, and the
interface lies in the x–y plane (normal to ẑ). Find the
electric field in medium 2 just below the interface.
Assume there is no free surface charge at the
boundary.
E₂ = 3x̂ + 1.6ẑ V/m
Tangential E is continuous: E₂ₓ = 3. Normal D is
continuous: ε₁E₁ₙ = ε₂E₂ₙ → 2(4) = 5E₂ₙ → E₂ₙ = 1.6.
