. . . B. ¹

The electric field at point A is in the positive x direction; the field at point B is in the negative x direction because the y components of the fields from the two charges cancel by symmetry, leaving only the x components, which are both negative.

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. . . zero? ²

The positive charge is closer, so its (positive) potential exceeds the negative potential of the negative charge.

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. . . B? ³

The positive and negative charges are equidistant from B, so their potentials cancel exactly.

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. . . perpendicular? ⁴

This question was on a WebCT Quiz. The electric field must be perpendicular to the velocity or the particle's speed would change, altering the magnetic force. The magnetic field need not be perpendicular to the velocity, as long as it has a perpendicular component big enough to cancel the electric force. And therefore the magnetic field need not be perpendicular to the electric field either.

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. . . ? ⁵

The EQUIPARTITION THEOREM states that in thermal equilibrium at temperature $\tau$, the rms energy associated with any "degree of freedom" for which positive and negative values yield the same energy is simply $\tau/2$. The charge on a capacitor is such a degree of freedom, since the energy stored in the capacitor is given by Q²/2C. Thus $\langle Q^2/2C \rangle = \langle Q^2 \rangle /2C = \tau/2$ or $\langle Q^2 \rangle = C \tau$. Here C = 1 F and $\tau = k_{\rm B} \times 300 = 0.4142 \times 10^{-20}$ J, giving $\langle Q^2 \rangle = 0.4142 \times 10^{-20}$ C² or $Q_{rms} = 0.6436 \times 10^{-10}$ C.

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. . . answer] ⁶

If the wires were parallel the capacitance would be infinite, but they are perpendicular, so the capacitance is finite. But calculating it would not be trivial.

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. . . Explain. ⁷

Like the usual lens coating, this one produces a phase shift of $\pi$ at the first reflection; but unlike the usual low-index coating, it does not produce any phase shift at the second reflection. So the two reflected rays will be out of phase for zero coating thickness - the ideal nonreflective coating! Now, of course one cannot make a coating of literally zero thickness, and a coating half a wavelength thick would also be nonreflective - for that wavelength! So the ideal thickness (nonreflective for all visible wavelengths) would be as close to zero as possible, or "much thinner than any of the wavelengths of interest".

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. . . credit.) ⁸

The temperature is defined as the inverse of the rate of change of entropy with energy. In general, this is independent of the absolute magnitude of either the energy or the entropy. Although there are some systems in which energy and entropy both increase with increasing temperature, this is not always true. Therefore only the next-to-last answer is correct. The last one would be right if one of the "increases" were a "decreases" but the derivative defines 1/T, not T.

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. . . loop? ⁹

Parallel currents attract; antiparallel currents repel. Thus a segment of wire on one side of the expanded loop repels one on the other side, causing the loop to expand. Another way of looking at it is that the coil contains a region of magnetic field which therefore has a positive energy density; spreading out to the largest diameter reduces the energy density, so there is a force "trying" to increase that diameter.

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. . . loop? ¹⁰

The magnetic field at the centre of a current loop is inversely proportional to the radius of the loop; this gives one factor of 1/2. There are also two turns in the double-loop coil, giving another factor of 1/2 for a total ratio of 1/4.

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. . . true? ¹¹

This question was on a WebCT Quiz. A simple application of Ampère's Law reveals that there can be no field inside the tube, since any Ampèrian loop in that region encloses no current; but the field outside the tube is (by symmetry) the same as if the current were carried by a wire down its axis.

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. . . line.) ¹²

Everyone should get this without help; it was provided partly as a "built in Summary Sheet" of all the fundamental relationships in Electricity & Magnetism.

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. . . field)? ¹³

Spin up means magnetic moment down; the potential energy of a magnetic moment in a magnetic field is given by $U = - \vec{\mbox{\boldmath$\mu$\unboldmath }} \cdot \vec{\mbox{\boldmath$B$\unboldmath }}$ so the spin up state is the higher energy state for the electron. The only way for the higher energy state to be preferentially populated is if the heat reservoir is at a vanishingly small negative temperature. (A reservoir of nuclear spins, for instance, could be prepared at such a temperature by suddenly reversing the magnetic field; but most "normal" reservoirs can only have positive temperatures, so an answer of "It's impossible!" is worth 2 out of 3 marks.)

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. . . field)? ¹⁴

This is the lower energy state and therefore its probability approaches 1 as the temperature approaches zero.

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. . . "down"? ¹⁵

This would be the situation for infinite temperature (positive or negative, they amount to the same thing). Note: this whole question could have been included in the "QUICKIES" section, as it involves no calculations at all.

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. . . closed. ¹⁶

The circuit must contain an inductance and a capacitance to oscillate. To get the prescribed frequency we need $\omega_0 \equiv \sqrt{1/LC} = 0.5$ s^-1, or LC = 4 s^-2. We can get an effective L of 2 H by connecting the two inductances in series and an effective C of 2 F by connecting the two capacitors in parallel:

0.75in

The resistors should be left out, since they would cause a shift in the actual frequency of oscillation.

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. . . =0. ¹⁷

The initial charge on each capacitor is Q₀ = 1 V $\times 1$ F = 1 C. The effective resistance is 4 $\Omega$, the effective capacitance is 1/2 F and the effective inductance is 2 H. You can quickly reproduce the derivation for k in the trial solution Q(t) = Q₀ e^kt:

$$k = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$$

where $\gamma \equiv R/2L = 4/(2\times2) = 1$ s^-1 and $\omega_0 \equiv \sqrt{1/LC} = \sqrt{1/(2 \times {1\over2}} = 1$ s^-1. Thus $k = - \gamma$ and the charge on each capacitor varies as . The current is just the derivative of Q with respect to time: . (The current is negative because the charge is decreasing.)

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. . . resistor. ¹⁸

The two rails, the resistor and the bar make a Faraday loop enclosing a magnetic flux $\Phi_M = \ell B (v t + x_0)$ where x₀ is the position of the bar at t=0, whenever that was. It doesn't matter, since to get the induced ${\cal{EMF}}$ we take the time derivative: ${\cal E}_{\rm ind} = - \dot{\Phi}_M = - \ell B v = (0.5$ m $) \times (0.675$ T $) \times 4.20$ m/s) or ${\cal E}_{\rm ind} = - 1.4175$ V. This voltage applied across the resistor gives a current of -0.1181 A, and the product of the voltage times the current (i.e. the power dissipated in the resistor) is therefore P = 0.1674 W.

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. . . .  ¹⁹

The first minimum of the diffraction pattern (defining the width of an individual slit) is at about 50 mm or 0.05 m, which at a distance of 1 m is equivalent to $\theta_1^{\rm diffr} \approx = \sin \theta_1^{\rm diffr} = 0.05$, so that the slit width is 1/0.05 = 20 times the wavelength: $a = 20 \; \mu$m. The fourth principal maximum of the intereference pattern appears to be right on top of the first diffraction mimimum, which means that the distance between slits is four times the slit width: $d = 80 \; \mu$m. Finally, there are four secondary maxima and five minima between principal maxima, so there are N = 6 slits.

2.75in

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