FE Electrical Electromagnetics Practice Problems

Answer: B) \(-1\)

Put \(Z_L = 0\) into the Handbook's definition:

$$\Gamma = \dfrac{Z_L - Z_0}{Z_L + Z_0} = \dfrac{0 - 75}{0 + 75} = -1$$

Everything is reflected, and inverted. The reason is the boundary condition itself: a short circuit cannot support any voltage, so the reflected wave must exactly cancel the incident one at the load, which requires equal amplitude and opposite sign. The standing wave ratio is infinite and no power reaches the load. The value of \(Z_0\) drops out - a short reflects with \(\Gamma = -1\) on any line.

+1 is the reflection from an open circuit, the other case of total reflection. There the current must vanish at the load instead of the voltage, so the voltage wave reflects without inversion.

0 is a matched load, \(Z_L = Z_0\), which absorbs everything and reflects nothing. A short is the opposite extreme.

−75 is \(Z_L - Z_0\) with the normalising denominator forgotten. A reflection coefficient is dimensionless and its magnitude never exceeds 1 for a passive load.

Answer: B) 0.75λ

The Handbook defines the wavelength as the distance the signal travels in one period, λ = U/f, with U the velocity of propagation. That velocity is the one the panel gives, not the speed of light:

$$\lambda = \dfrac{v_p}{f} = \dfrac{2 \times 10^{8}}{500 \times 10^{6}} = 0.4\ \text{m}$$

$$\dfrac{l}{\lambda} = \dfrac{0.3}{0.4} = 0.75$$

so the line is three quarters of a wavelength long. In the electrical terms the same page uses, β = 2π/λ and the electrical length is βl = 2π(0.75) = 1.5π radians, or 270°.

The characteristic impedance is in the figure and plays no part here. It matters for what the line does to a load, not for how many wavelengths fit on it, and a question that gives a number the answer does not use is testing whether you know which numbers belong.

0.50λ is the answer with the speed of light used instead of the phase velocity: 3 × 10⁸ over 500 MHz gives 0.6 m and 0.3/0.6 = 0.5. On a real line the signal is slower than light, and how much slower is what the dielectric between the conductors decides, so a phase velocity is always given when it matters. 1.33λ is the ratio the other way up, λ/l rather than l/λ; the line is shorter than a wavelength and a half, so an answer above 1 should prompt a second look at which quantity was divided by which. 1.50λ takes the 1.5 out of βl = 1.5π radians and reports it as a count of wavelengths, but a full wavelength is 2π radians, not π.

Answer: C) 3.14 × 10⁻⁵ T

Substitute and let the π survive:

$$B = \dfrac{\mu_0 I}{2R} = \dfrac{(4\pi \times 10^{-7})(5)}{2(0.1)} = \dfrac{20\pi \times 10^{-7}}{0.2} = \pi \times 10^{-5} = 3.14 \times 10^{-5}\ \text{T}$$

The answer is π × 10⁻⁵ exactly, which is a good sign: in this family of results the 4π of μ₀ usually cancels against a 2π from the geometry, and here it does not, because the loop's geometry contributes a 2 rather than a 2π. That single difference is where the first wrong answer comes from.

1.00 × 10⁻⁵ T is what the straight-wire expression gives, μ₀I/(2πR), and it is the one formula of the two that the Handbook actually prints. It is π times smaller than the right answer, and the difference is physical rather than arithmetic: an infinite straight wire spreads its field over the whole circumference 2πR, while every element of a closed loop contributes to the field at the centre with the same sign, so the loop concentrates it. Reaching for the wire formula because it is the one in the book is the trap here.

1.57 × 10⁻⁵ T is μ₀I/(4R), which is the field at the centre of a semicircular arc: half the turn, half the field. 6.28 × 10⁻⁵ T is μ₀I/R, the 2 in the denominator dropped.

Answer: A) \(20\,\text{mm}\)

The Handbook defines the wavelength as \(\lambda = U/f\), the distance the signal travels in one period, with \(U\) the velocity of propagation. In free space \(U = c\) and

$$\lambda_0 = \dfrac{c}{f} = \dfrac{3 \times 10^8}{10 \times 10^9} = 0.03\text{ m} = 30\text{ mm}$$

Inside the dielectric the wave is slowed to \(U = c/\sqrt{\mu_r\varepsilon_r} = c/1.5\), while the frequency - set by the source - is unchanged, so the wavelength shrinks by the same factor:

$$\lambda = \dfrac{\lambda_0}{\sqrt{\varepsilon_r}} = \dfrac{30}{1.5} = 20\text{ mm}$$

Frequency is the invariant across a boundary; velocity and wavelength are what change. A wave entering glass keeps its colour but not its wavelength.

30 mm is the free-space wavelength with the medium ignored.

45 mm multiplies by \(\sqrt{\varepsilon_r}\) instead of dividing. A wave cannot be longer inside a dielectric than in free space, because it cannot travel faster there.

67.5 mm multiplies by \(\varepsilon_r\) itself, compounding the direction error with a missing square root.

Answer: C) 4520 V/m

The Handbook gives the field of one infinite sheet, and it is the same everywhere on either side of it:

$$E_{\text{one sheet}} = \dfrac{\rho_s}{2\varepsilon_0} = \dfrac{40 \times 10^{-9}}{2(8.85 \times 10^{-12})} = 2260\ \text{V/m}$$

Now superpose. Between the sheets, the positive one pushes downward and the negative one pulls downward, so the two contributions point the same way and add:

$$E = \dfrac{\rho_s}{2\varepsilon_0} + \dfrac{\rho_s}{2\varepsilon_0} = \dfrac{\rho_s}{\varepsilon_0} = 4520\ \text{V/m}$$

Outside the pair they point opposite ways instead, and since they are equal in magnitude they cancel exactly. That is the whole of the parallel-plate capacitor in two lines: a uniform field between, nothing outside.

The separation is given and is not used, and that is deliberate. The field of an infinite sheet does not depend on how far away you are, so the field between the pair does not depend on d either. What d does control is the voltage across the gap, V = Ed = 45.2 V here, and the capacitance per unit area. If moving the plates apart changed the field, a capacitor's charge would not stay put when you pulled its plates.

2260 V/m is one sheet counted and the other forgotten. 9040 V/m doubles a second time, adding two full ρₛ/ε₀ contributions instead of two halves. 0 V/m is the correct answer for the region outside the pair, and picking it means reading the wrong region of the figure.