. . . . ¹

This is partly an exercise in juggling units, but mainly it requires remembering that ${\displaystyle \epsilon_{_0} \mu_{_0} = {1 \over c^2}}$ so that ${\displaystyle \epsilon_{_0} = {1 \over \mu_{_0} c^2} = {1 \over 4\pi \times 10^{-7} \times (2.9979 \times 10^8)^2}}$ or $$\epsilon_{{_0}}^{}$$=0.8854×10^-11 C²/N ^. m²  .

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. . . coil?²

You must first recognize that this is a thermal physics problem involving the equipartition theorem: $\langle$$\varepsilon$$\rangle$ = ${1\over2}$k_BT. Here $\varepsilon$ = ${1\over2}$LI² is the energy stored in a coil with inductance L carrying current I. So we have in one step ${1\over2}$L$\langle$I²$\rangle$ = ${1\over2}$k_BT or ${\displaystyle \sqrt{\langle I^2 \rangle} = \sqrt{k_{_{\rm B}} T \over L} = \sqrt{1.3807 \times 10^{-23} \times 300 \over 1}}$ or I_rms = 6.43×10^-11 A  .

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. . . contribution?³

Maxwell added the displacement current ${\displaystyle \int\!\!\!\!\int \left( \partial \vec{\mbox{\boldmath$D$\unboldmath}} \over \partial t \right) \cdot d\vec{\mbox{\boldmath$A$\unboldmath}}}$ to the right side of Ampère's Law, $$\oint$$$$\vec{{\mbox{\boldmath$H$\unboldmath}}}\,$$ ^. d$$\vec{{\mbox{\boldmath$\ell$\unboldmath}}}\,$$=$$\int$$$$\int$$$$\vec{{\mbox{\boldmath$J$\unboldmath}}}\,$$ ^. d$$\vec{{\mbox{\boldmath$A$\unboldmath}}}\,$$. (Differential forms are also OK, as are versions with $\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$ and $\mu$ or $\vec{{\mbox{\boldmath$E$\unboldmath}}}\,$ and $\epsilon$. Just saying the words was worth full credit.)

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. . . \dots⁴

Let x be the scale factor. We know $\tau$ = L/R so all we need is to deduce the effect on L and R of x$\to$2x. We have L = NAB/I = $\pi$r²$\mu_{0}^{}$N²/$\ell$ $\propto$ r²/$\ell$ $\propto$ x²/x = x, so L$\to$2L as x$\to$2x. Meanwhile R = $\rho$$\ell$/A $\propto$ x/x² = 1/x so R$\to$R/2 as x$\to$2x. Together these give $\tau$$\to$4$\tau$ (i.e. $\tau$ is quadrupled.

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. . . possible.⁵

Thin film interference. This was meant as a gift. Details are not required, but the basic idea is that a film of the right thickness will produce constructive interference between rays reflected off the front and back sides of the film if the wavelength is red, and destructive interference for other wavelengths. Thicker films are more selective, but harder to grow with sufficient uniformity.

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. . . space.)⁶

Something like this:

The initial acceleration is in the + y direction due to $\vec{{\mbox{\boldmath$F$\unboldmath}}}_{{\rm E}}^{}$ = q$\vec{{\mbox{\boldmath$E$\unboldmath}}}\,$. Once the proton is moving, $\vec{{\mbox{\boldmath$F$\unboldmath}}}_{{\rm M}}^{}$ = q$\vec{{\mbox{\boldmath$v$\unboldmath}}}\,$×$\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$ sets in. Eventually the path must go "horizontal" (in the + x direction) and then turn around, slow back down due to $\vec{{\mbox{\boldmath$F$\unboldmath}}}_{{\rm E}}^{}$ and come back to rest at the same y₀ where it started, but displaced sideways in x. Then the process starts over. This configuration produces a gradual sideways drift in the + x direction; it is used in many types of particle detectors.

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. . . reasoning.⁷

Obviously we are going to use Ampère's Law, $$\oint$$$$\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$$ ^. d$$\vec{{\mbox{\boldmath$\ell$\unboldmath}}}\,$$=$$\mu_{0}^{}$$I_encl, but first we need to make some observations about symmetry: as long as we stay far away from the ends of the sheet, it "looks infinite" and there is no way to choose any preferred value of y. Thus $\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$ must be independent of y. When combined with the right hand rule for circulation of $\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$ around the current through the rectangular Ampèrian loop shown below, this implies that $\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$ must be along the ±y direction as shown, and uniform in magnitude. This makes the line integral trivial: it is just 2B$\ell$. The enclosed current is then J times the enclosed area, or J$\ell$d, giving 2B$\ell$ = $\mu_{0}^{}$J$\ell$d or B = ${1\over2}$$\mu_{0}^{}$Jd in the directions shown, independent of y or x (except insofar as it is up on the right and down on the left. (Such a device is called a septum magnet. It is used to split a beam into two parts, both deflected away from the current sheet.)

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. . . sentence.]⁸

The net charge on the ball is zero, but it is a conductor and the charges can move. Negative charges will move toward the positive external charge and pile up on that side, while an equal amount of positive charge will move toward the far side and pile up there. This is known as polarizability. In this configuration, the positive charges in the ball are further away from the external charge than the negative charges, and so feel a weaker force; as a result the ball feels a net attraction to the external charge and will swing toward it.

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. . . it.⁹

Like this:

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. . . resistor.¹⁰

The resonant frequency is $\omega_{0}^{}$ = 1/$\sqrt{{LC}}$ = 1/$\sqrt{{3 \times 10^{-5} \times 0.2345}}$ = 377 s^-1. Remember that Hz are cycles per second; to convert to an angular frequency we must multiply 60 Hz by 2$\pi$, giving $\omega$ = 2$\pi$×60 = 377 s^-1. So we are indeed "on resonance". At this frequency both L and C "disappear" and the circuit acts as if we were simply running the driving voltage through the resistor by itself - in which case I = $\cal {E}$/R and the amplitude I₀ of the current oscillations is just 15 V divided by 30 $\Omega$ or I₀ = 0.5 A  .

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. . . calibration?¹¹

Units of inverse energy are the natural units for inverse temperature ${\displaystyle \beta \equiv {\partial \sigma \over \partial U}}$ where $\sigma$ $\equiv$ ln$\Omega$ is the dimensionless entropy and U is the energy of the system. Energy will flow spontaneously from regions of low $\beta$ to regions of high $\beta$, so lower $\beta$ means hotter. Unlike its inverse, the temperature $\tau$ $\equiv$ 1/$\beta$, this quantity does not exhibit any divergences where the derivative goes to zero, and is therefore a more "rational" characterization of "hotness".

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. . . required!]¹²

Judging from the calibration marks, hotter (lower $\beta$) is to the right. Thermal expansion increases as we get hotter, so whatever liquid is expanding must push the interface away from the bulb as it gets hotter. This places the bulb on the left. (The liquid is clear and the red colour on the empty side is presumably due to some trick of refraction that is spoiled by the liquid.)

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. . . Kelvin?¹³

The thermometer reads 40 eV ^-1 = $\beta$ = 1/$\tau$ = 1/k_BT so ${\displaystyle T = {1 \over k_B} \cdot {1 \; {\rm eV} \over 40} = {1.602 \times 10^{-19} \; {\rm J} \over 40 \times 1.3807 \times 10^{-23} \; {\rm J/K} }}$ or T = 290.11 K  . A nice comfortable room temperature.

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. . . conductor;¹⁴

Zero, as always. This was meant as a gift.

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. . . cavity;¹⁵

By Gauss' Law with spherical symmetry, field lines originate on + Q at the centre and drop off radially as 1/$\rho^{2}_{}$ where $\rho$ is the distance from the centre of the cavity. (You can use any symbol you like for this distance, but not the same one as the distance from the axis of the cylinder.) When they reach the inner surface of the cavity, field lines are terminated by induced negative charges so that none penetrate into the conductor. Thus ${\displaystyle \vec{\mbox{\boldmath$E$\unboldmath}}(\rho) = {Q \over 4 \pi \epsilon_0} {\hat{\mbox{\boldmath$\rho$\unboldmath}} \over \rho^2}}$  .

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. . . cylinder.¹⁶

First you must realize that the positive charges left behind when - Q is induced on the inner surface of the cavity will try to get as far away from each other as possible. They know nothing of the distribution of charges inside the conductor; all that is shielded by the conductor. Thus a positive charge + Q is evenly distributed over the outside of the cylinder. By Gauss' Law with cylindrical symmetry, this gives the same field outside the cylinder as a long line of charge with uniform charge per unit length $\lambda$, namely ${\displaystyle E = {\lambda \over 2 \pi \epsilon_0 \; r}}$. Finally you must realize that $\lambda$ = Q/$\ell$, giving ${\displaystyle \vec{\mbox{\boldmath$E$\unboldmath}}(r > R) = {Q \over 2 \pi \epsilon_0 \ell \; r} \; \hat{\mbox{\boldmath$r$\unboldmath}}}$  .

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. . . deflection?¹⁷

$\vec{{\mbox{\boldmath$F$\unboldmath}}}\,$ = q($\vec{{\mbox{\boldmath$E$\unboldmath}}}\,$ + $\vec{{\mbox{\boldmath$v$\unboldmath}}}\,$×$\vec{{\mbox{\boldmath$B$\unboldmath}}}\,$) = 0 if E = vB in this geometry. Thus the beam passes without deflection when E = 10⁶×0.01 = 10⁴ V/m  .

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. . . field?¹⁸

Square plates with area 1 m² implies a side length $\ell$ = 1 m. At 10⁶ m/s this takes 10^-6 s = 1 $\mu$s  . This was partly a gift and partly to avoid confusion about whether the protons can make it through the separator undeflected in a time very short compared to the time it takes for the electric field to change appreciably in the next question.

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. . . plates.¹⁹

The electric field between the plates is ${\displaystyle E = {\sigma \over \epsilon_0} = {Q \over \epsilon_0 A}}$ so E $\propto$ Q and initially ${\displaystyle E_0 = {Q_0 \over \epsilon_0 A} = {10^{-6} \over 0.8854 \times 10^{-11} \times 1} = 1.1294 \times 10^5}$ V/m. For an RC circuit Q(t) = Q₀e^-t/$\tau$ where $\tau$ = RC. Since C = Q/V and V = Ed, we have ${\displaystyle C = {Q \over d} \cdot {\epsilon_0 A \over Q} = {\epsilon_0 A \over d} = {0.8854 \times 10^{-11} \times 1 \over 0.01} = 0.8854 \times 10^{-9}}$ F, giving $\tau$ = RC = 0.8854×10^-9 F  ×10⁸  $\Omega$ = 0.08854 s. We want t such that ${\displaystyle {E \over E_0} = {10^4 \over 1.1294 \times 10^5} = 0.08854 = e^{-t/0.08854}}$ or ln(0.08854) = - t/0.08854 or t = 0.21465 s  .

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. . . negative.²⁰

There are at least two valid ways to solve this problem.

One way is to set up a wedge-shaped closed path that goes out to the ball along the original position of the wire, along the path followed by the ball, and back to the centre pivot along the present wire position; then apply Faraday's Law to the magnetic flux through the path and find the induced EMF around the loop (which is the voltage across the wire) at a given instant. The "pie wedge" has an area A which is in the same ratio to $\pi$R² as $\theta$ is to 2$\pi$: A/$\pi$R² = $\theta$/2$\pi$ (where $\theta$ = $\omega$t is the angle between the original position of the wire and its current position). The magnetic flux enclosed by this loop is thus $\Phi_{M}^{}$ = B$\pi$R²$\omega$t/2$\pi$ = ${1\over2}$$\omega$R²Bt and its time derivative is just ${1\over2}$$\omega$R²B. Thus $\cal {E}$_ind = - ${1\over2}$$\omega$R²B = - ${1\over2}$30(0.4)²(0.15) =  -0.36 V  . This voltage acts between the ball and the pivot, the only part of the Faraday loop that is not "virtual". The downward (into the page) flux through the loop is increasing as the area of the loop increases, so by Lenz's Law the induced EMF is in the direction that would produce a current (if it could) whose magnetic field would counteract the change in flux - namely a field upward (out of the page), which would be generated by a counterclockwise current, which means the ball is positive and the pivot is negative.

The second approach is to consider the Hall effect on charges in the wire: a small element of the wire with length dr at position r < R is moving at a speed v = $\omega$r through the field, producing a force F = qvB = q$\omega$rB which is inward (toward the pivot) for a positive charge and outward (toward the ball) for a negative charge. Thus charges will move in those directions and build up on the ball or pivot end until an electric field E is set up (by the charge imbalance) that just cancels the magnetic force: qE = q$\omega$rB or E = $\omega$Br. The voltage between pivot and ball is the integral of this electric field along the path: V = - $\int_{0}^{R}$E(r)dr = - $\omega$B$\int_{0}^{R}$rdr = - ${1\over2}$$\omega$BR² = - ${1\over2}$×30×0.15×(0.4)² =  -0.36 V  , in agreement with our earlier result. The only trouble is, this version predicts a positively charged pivot and a negatively charged ball - the opposite of the previous result.

Which is right?

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. . . ?²¹

The angle $\vartheta_{m}^{{\rm I}}$ of the m^th principal maximum of the N-slit interference pattern is where the phase difference between two adjacent slits is m(2$\pi$) and/or the corresponding path difference is m$\lambda$.

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. . . ?²²

The angle $\vartheta_{1}^{{\rm D}}$ of the first minimum of the single-slit diffraction envelope is where the phasors from N "pseudoslits" making up the slit width all add up to form a closed N-sided regular polygon with N$\to$$\infty$.

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. . . ?²³

The angular separation $\Delta$$\vartheta^{{\rm I}}_{}$ between any principal maximum (PM) of the N-slit interference pattern and the first minimum adjacent to that PM is where the phasors from the N slits add up to form a closed N-sided regular polygon.

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. . . ? ²⁴

The width $\Delta$$\vartheta^{{\rm I}}_{}$ gets smaller with larger N (see previous formula), giving sharper principal maxima, which is especially important when the grating is used to resolve different colours. (See Dispersion and Resolving Power.

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. . . maxima? ²⁵

  N - 1  .

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