Time Varying Fields and Maxwells Equations Questions and Answers

Answer: C

Under time reversal \(\vec{E} \to \vec{E}\) and \(\vec{B} \to -\vec{B}\).

Charges do not change under \(t \to -t\), and the electric field is fixed by where the charges are, so \(\vec{E}\) is unchanged. Currents are charges in motion; reversing time reverses every velocity, so every current reverses, and the magnetic field produced by currents changes sign.

The Lorentz force confirms this: \(\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})\) is even under time reversal (it is mass times acceleration), and \(\vec{v}\) is odd, so \(\vec{B}\) must be odd while \(\vec{E}\) is even. The other three combinations break this invariance.

Answer: C

The correct graph is (C).

The side of length l is perpendicular to the motion, so while the loop is entering or leaving the field the flux changes at the rate \(Blv\), and the emf has magnitude \(Blv\). This rules out (A) and (B), which use \(Bwv\).

• From \(x = 0\) to \(w\) the loop enters and the inward flux grows, so \(\varepsilon = -d\Phi/dt = -Blv\).
• From \(w\) to \(4w\) the loop is wholly inside and the flux is constant, so \(\varepsilon = 0\).
• From \(4w\) to \(5w\) it leaves and the flux falls, so \(\varepsilon = +Blv\).

With inward flux counted positive, the pulse on entry is negative and on exit positive, as in (C); (D) has the signs reversed.

Answer: B

The displacement current at P is inversely proportional to r and tangential.

The changing flux inside the solenoid (radius a) induces a circulating electric field. Outside it, Faraday's law on a circle of radius r gives

\[ \begin{aligned} E \cdot 2\pi r &= -\frac{d}{dt}(B_{0}t^{2}\pi a^{2}) \\ E &= -\frac{B_{0} t\, a^{2}}{r} \end{aligned} \]

The displacement current density is \(\varepsilon_{0}\,\partial \vec{E}/\partial t = -\varepsilon_{0} B_{0} a^{2}/r\) along \(\hat{\phi}\): tangential, constant in time, and proportional to \(1/r\).

It is not radial, because the induced field circulates. It does not grow with time, because \(E\) is linear in \(t\) so its derivative is constant. The \(1/r^{2}\) option has the wrong dependence for a field produced by a fixed enclosed flux.

Answer: C

The correct statement is that the emf induced in loop P is \(\propto \frac{1}{s^4}\).

Because \(s\) is much larger than both radii, loop Q acts as a magnetic dipole \(m = I\pi R_2^2\), and its field on the axis at P falls off as the cube of the distance:

\[ \begin{aligned} B &= \frac{\mu_0 m}{2\pi s^{3}}, \quad \Phi_P \approx B\,\pi R_1^{2} \propto \frac{1}{s^{3}} \\ \varepsilon &= -\frac{d\Phi_P}{dt} = -\frac{d\Phi_P}{ds}\frac{ds}{dt} \propto \frac{v}{s^{4}} \end{aligned} \]

So the emf is proportional to \(1/s^4\) and to \(v\) to the first power.

The field at the centre of P goes as \(1/s^3\), not \(1/s^2\), and so does the flux, which rules out the first two statements. The emf is linear in \(v\), not proportional to \(v^2\).

Answer: A

The correct matching is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i).

• Ohm's law at a point relates current density to field: J = σE, (iv).
• Kirchhoff's voltage law says the emfs round a loop equal the IR drops: ΣV = IΣR, (iii).
• Kirchhoff's current law in point form, for steady currents, is ∇·J = 0: no charge accumulates, (ii).
• Gauss's law at a point is ∇·D = ρ, (i).

The other options swap the two Kirchhoff laws or pair Ohm's law with Gauss's law.

Answer: D

The amplitude is 18 A/m².

The displacement current density is the time derivative of the electric flux density:

\[ \begin{aligned} J_{d} &= \frac{\partial D}{\partial t} \\ &= 3 \times 10^{-7} \times 6 \times 10^{7} \cos(\dots) \\ &= 18\cos(6 \times 10^{7}t - 0.35x)\ \text{A/m}^{2} \end{aligned} \]

The amplitude is 18 A/m². The relative permittivity is not needed, because D is given directly; \(\epsilon_{r}\) would matter only if the field were given as E. The smaller options come from dropping or mis-multiplying the factors 3 and 6.

Answer: C

In free space the equation becomes \(\nabla \times \bar{H} = \dfrac{\partial \bar{D}}{\partial t}\).

Free space has no conductors and no free charges, so the conduction current density \(\bar{J}_{cond} = \sigma\bar{E}\) is zero (\(\sigma = 0\)). Only the displacement current term remains.

Setting the whole right side to zero would also drop the displacement current, which is what lets electromagnetic waves propagate through free space. Keeping only \(\bar{J}_{cond}\) is the form for a good conductor at low frequency, the opposite case, and 'infinite' has no physical meaning here.

Answer: C

The correct matching is a-ii, b-iv, c-iii, d-i.

• Ampere's law, with Maxwell's displacement current: \(\nabla \times \bar{H} = \bar{J} + \partial\bar{D}/\partial t\) (ii).
• Faraday's law: \(\nabla \times \bar{E} = -\partial\bar{B}/\partial t\) (iv).
• Gauss' law for electric fields: \(\nabla \cdot \bar{D} = \rho\) (iii).
• No magnetic monopoles: \(\nabla \cdot \bar{B} = 0\) (i).

The curl equations describe circulation (Ampere and Faraday) and the divergence equations describe sources (charge, and the absence of magnetic charge). The near-miss code swaps the two divergence equations, giving the monopole law the charge density.

Answer: B

The induced emf is \(\pi r^{2}\alpha B_{0} e^{-\alpha t}\).

The field is along z and the loop lies in the xy plane, so the flux is field times area:

\[ \begin{aligned} \Phi &= \pi r^{2} B_{0} e^{-\alpha t} \\ \varepsilon &= -\frac{d\Phi}{dt} = \pi r^{2}\alpha B_{0} e^{-\alpha t} \end{aligned} \]

The emf is not zero because the flux changes with time. The options with \(2\pi r\) use the circumference where Faraday's law needs the enclosed area; one of them also omits the factor \(\alpha\) that comes from differentiating the exponential.

Answer: B

The Lorentz gauge condition is \(\vec{\nabla}\cdot\vec{A}\) \(+ \mu_{0}\varepsilon_{0}\,\partial V/\partial t = 0\).

The Lorentz (Lorenz) gauge chooses the potentials so that the divergence of \(\vec{A}\) cancels the time derivative of V divided by \(c^{2}\). With \(c^{2} = 1/\mu_{0}\varepsilon_{0}\):

\[ \begin{aligned} &\vec{\nabla}\cdot\vec{A} \\ &+ \frac{1}{c^{2}}\frac{\partial V}{\partial t} = 0 \end{aligned} \]

In this gauge both V and \(\vec{A}\) satisfy inhomogeneous wave equations, and the condition is Lorentz invariant. Writing \(1/\mu_{0}\varepsilon_{0}\) instead multiplies by \(c^{2}\) where it should divide, the minus sign gives the wrong relative sign, and the second time derivative belongs to the wave equation, not the gauge condition.