FE Electrical Electromagnetics Practice Problems

Answer: B) \(3.37\,\text{N}\)

Coulomb's law in the Handbook's form uses the permittivity directly, \(F = Q_1Q_2/(4\pi\varepsilon r^2)\) with \(\varepsilon_0 = 8.85 \times 10^{-12}\) F/m. The constant \(1/(4\pi\varepsilon_0)\) is \(8.99 \times 10^9\) N·m²/C², and only the magnitudes of the charges enter the magnitude of the force:

$$F = \dfrac{(8.99 \times 10^9)(3 \times 10^{-6})(5 \times 10^{-6})}{(0.2)^2} = \dfrac{(8.99 \times 10^9)(1.5 \times 10^{-11})}{0.04} = 3.37\text{ N}$$

The opposite signs make the force attractive, which is a statement about direction and does not change the magnitude. Note the order of magnitude: microcoulombs squared is \(10^{-12}\), and that exponent is where most slips on this item happen.

0.674 N divides by \(r\) instead of \(r^2\). Coulomb's law is an inverse-square law; dividing by 0.2 once leaves the answer a factor of 5 too small.

33.7 N is a power-of-ten slip in the charges, treating \(\mu\text{C} \times \mu\text{C}\) as \(10^{-11}\) rather than \(10^{-12}\).

42.4 N drops the \(4\pi\) and writes \(F = Q_1Q_2/(\varepsilon_0 r^2)\). The Handbook's form carries the \(4\pi\) explicitly; if you prefer \(k = 8.99 \times 10^9\), remember that it already contains it.

Answer: C) \(0.08\,\text{V}\)

Faraday's law in the Handbook's form is \(v = -N\,d\phi/dt\). With one turn and the field perpendicular to the loop, \(\phi = BA\), and because the rise is uniform the derivative is just the ratio of the changes:

$$|v| = A\,\dfrac{\Delta B}{\Delta t} = (0.04)\,\dfrac{0.5 - 0.1}{0.2} = (0.04)(2.0) = 0.08\text{ V}$$

Only the rate of change of flux matters. A loop sitting in a steady 0.5 T field has no emf at all; this one has 80 mV only while the field is changing, and the sign, by Lenz's law, is such that the induced current opposes the rise.

0.016 V is \(A\,\Delta B\), the change in flux in webers, with the time never applied. Flux is not emf; its rate of change is.

0.1 V uses the final field, \(A B_2/\Delta t\), in place of the change \(A(B_2 - B_1)/\Delta t\). The 0.1 T that was already there induces nothing.

0.02 V makes the opposite slip and uses the initial field alone.

Answer: D) \(4.0\)

The Handbook gives the standing wave ratio directly in terms of the reflection coefficient's magnitude:

$$\text{SWR} = \dfrac{1 + |\Gamma|}{1 - |\Gamma|} = \dfrac{1 + 0.6}{1 - 0.6} = \dfrac{1.6}{0.4} = 4.0$$

The ratio is exactly what its name says: where the incident and reflected waves add, the voltage envelope peaks at \(1 + |\Gamma|\) times the incident amplitude, and where they cancel it dips to \(1 - |\Gamma|\). An SWR of 4 is a serious mismatch - 36% of the incident power is being reflected. The 75 \(\Omega\) characteristic impedance is not needed; SWR depends on \(|\Gamma|\) alone, whatever line it happens to be on.

1.67 is \(1/|\Gamma|\), and 2.5 is \(1/(1 - |\Gamma|)\): each is a fragment of the formula standing in for the whole.

0.25 is the formula inverted, \((1 - |\Gamma|)/(1 + |\Gamma|)\). An SWR is never less than 1, since the envelope maximum cannot be smaller than the minimum, so this one can be rejected on sight.

Answer: B) \(4.0 \times 10^{-5}\,\text{T}\)

The Handbook gives the field of a current-carrying wire on the \(z\)-axis as \(H = B/\mu = I/(2\pi r)\,a_\phi\), so

$$B = \dfrac{\mu_0 I}{2\pi r} = \dfrac{(4\pi \times 10^{-7})(10)}{2\pi(0.05)} = \dfrac{2 \times 10^{-6}}{0.05} = 4.0 \times 10^{-5}\text{ T}$$

That is 40 µT, roughly the strength of the Earth's field, which is a useful sanity check for any wire carrying a few amperes at a few centimetres. The direction is circumferential, given by the right-hand rule with the thumb along the current.

2.0 × 10⁻⁵ T uses \(4\pi r\) in the denominator, borrowing the constant from the Biot-Savart integrand instead of its result for an infinite wire. The Handbook writes the wire result with \(2\pi r\).

8.0 × 10⁻⁵ T drops the 2, writing \(\mu_0 I/(\pi r)\).

1.26 × 10⁻⁴ T is \(\mu_0 I/(2r)\), the field at the centre of a circular loop of radius \(r\). The loop formula has no \(\pi\) in the denominator because the whole circumference contributes; a straight wire's field falls off with the \(2\pi r\) of the Amperian path around it.

Answer: C) \(11\)

The divergence is the dot product of the del operator with the field, so each component is differentiated with respect to its own coordinate and the results are added:

$$\nabla \cdot \vec{F} = \dfrac{\partial(xy)}{\partial x} + \dfrac{\partial(2yz)}{\partial y} + \dfrac{\partial(3xz)}{\partial z} = y + 2z + 3x$$

$$\nabla \cdot \vec{F}\,\big|_{(1,2,3)} = 2 + 6 + 3 = 11$$

A positive divergence means the point is a net source of the field - more flux leaves a small volume around it than enters. In electrostatics that is Gauss's law in differential form, \(\nabla \cdot D = \rho\), which is why the operation matters here.

14 differentiates each component with respect to the other variable in it: \(\partial(xy)/\partial y + \partial(2yz)/\partial z + \partial(3xz)/\partial x = x + 2y + 3z = 1 + 4 + 9\). The divergence pairs \(F_x\) with \(x\), \(F_y\) with \(y\) and \(F_z\) with \(z\); the cross-pairings belong to the curl.

9 loses the coefficient 3 in the last term, giving \(2 + 6 + 1\).

8 loses the coefficient 2 in the middle term, giving \(2 + 3 + 3\).