. . . either;¹

   

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. . . either;²

   

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. . . either;³

   

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. . . either.⁴

  This is impossible, of course. Trust your own conclusions!

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. . . particle.⁵

  The Lorentz force on a moving charge is $\vec{\mbox{\boldmath$F$\unboldmath}} = q \vec{\mbox{\boldmath$v$\unboldmath}} \times \vec{\mbox{\boldmath$B$\unboldmath}}$, always perpendicular to $\vec{\mbox{\boldmath$v$\unboldmath}}$. Thereofre it can only change the direction of $\vec{\mbox{\boldmath$v$\unboldmath}}$, never its magnitude, and so ${1\over2}mv^2$ is constant.

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. . .  C: ⁶

  The charge on the capacitor bleeds off through the resistor exponentially: $Q(t) = Q_0 \exp(-t/\tau_{\rm RC})$, where $Q_0 - 1$ C and $\tau_{\rm RC} = RC = 1$ s.

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. . . ⁷

  The current in the inductance builds up from zero, asymptotically approaching the value it would have without the inductance present, namely $I_\infty = {\cal E}/R = 1$ A. $I(t) = I_\infty \left[1 - \exp(-t/\tau_{\rm LR})\right]$, where $\tau_{\rm LR} = L/R = 10^{-3}$ s = 1 ms.

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. . . Why?⁸

  The increasing upward flux through the loop induces an EMF around the loop in the direction that will generate its own downward field through the loop - i.e. clockwise. The force on the induced current due to the external field is toward the centre of the loop on both sides, causing it to collapse inward. Since the loop is a flattened oval, this makes it tend to flatten more.

On the other hand, the fields generated by the induced current act repulsively on the opposite side of the loop where the current is going in the opposite direction (opposite currents repel), so we expect the collapse of the loop to stop when the two sides get too close together.

If the loop is superconducting, it is not obvious which force "wins" the battle between attraction and repulsion. This is a tantalizing problem and gave you a chance to show off the subtlety of your thinking. Full credit will be given for any answer that makes sense.

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. . . words.)⁹

  Inside the wire, $\vec{\mbox{\boldmath$B$\unboldmath}}$ loops around the axis in the sense of the right hand rule for the inner current. Outside, $B=0$.

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. . . cable.¹⁰

  Outside, $B=0$ by Ampère's Law because the net current enclosed by a loop around the cable is zero.

Inside, Ampère's Law gives $\oint \vec{\mbox{\boldmath$B$\unboldmath}}\cdot d\vec{\mbox{\boldmath$\ell$\unb . . . . . . = \mu_o I_{\rm encl} = \mu_o I_0 \left(\not{\pi}r^2 \over \not{\pi}a^2\right)$ or ${\displaystyle B(r<a) = {\mu_o I_0 \over 2\pi a^2} \cdot r}$ .

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. . . . ¹¹

  The magnetic field is completely contained inside the wire; it makes loops around the axis and its strength varies with $r$. So we make an area element $dA = \ell dr$ consisting of a long strip of width $dr$ at radius $r$, oriented perpendicular to the field. The magnetic flux through this strip is $d\Phi = B(r) dA = {\mu_o I_0 \over 2\pi a^2} \ell r dr$ which integrates easily to give $\Phi = {\mu_o I_0 \over 2\pi a^2} \ell {1\over2}a^2 = \left( \mu_o \ell \over 4\pi \right) I_0$. The definition of $L$ is $\Phi = L I$, so ${\displaystyle L = {\mu_o \ell \over 4\pi}}$ . The only hard part here is visualizing the geometry for the flux integral.

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. . . switch.¹²

  If a battery was required to "drive" the current in the first place, then the cable must have a resistance $R$, in which case it constitutes an $LR$ circuit and has a time constant $\tau_{\rm LR} = L/R$ for the exponential decay of the current: $I(t) = I_0 \exp(-t/\tau_{\rm LR})$.

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. . . cable?¹³

  The definition of resistivity can be expressed as $R = \rho \ell/A$ where $A = \pi a^2$. Thus ${\displaystyle R = {10^{-6} \times 2 \over \pi \times (10^{-3})^2} }$ or ${\displaystyle R = {2 \over \pi} = 0.6366}$ $\Omega$ .

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. . . conductor?¹⁴

  You can express Ohm's Law either as $\vec{\mbox{\boldmath$E$\unboldmath}} = \rho \vec{\mbox{\boldmath$J$\unboldmath}}$, in which case ${\displaystyle E = \rho \, {I_0 \over \pi a^2}}$, or as $V = I R$ where $V = E \ell$, in which case ${\displaystyle E = {I_0 R \over \ell} = {0.5 \times 2/\pi \over 2}}$ or ${\displaystyle E = {0.5 \over \pi} = 0.159}$ V/m .

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