Self-check: 203-SN2-RE, Electricity and Magnetism

a) The smallest division is $1$ mm, so $\Delta L=0.5$ mm $=0.05$ cm. Each end of the object is read to within half a division.

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a) Threshold of slipping: $f_{s,max}=\mu_{s}N=\mu_{s}mg=(0.50)(20)(9.8)=98$ N. Since 90 N $<$ 98 N, the crate does not move. The friction then equals exactly 90 N (not 98 N!): static friction adjusts itself to cancel the net force, and reaches its maximum only at the threshold of slipping.

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b) The datasheet gives two terms to ADD. Proportional term: $0.5\,\%\times 12.47=0.0624$ V. Digit term: $2\times 0.01=0.02$ V. Total: $\Delta U=0.0624+0.02=0.0824$ V, which is rounded to ONE significant figure, $0.08$ V. The value is aligned on the same decimal: $U=(12.47\pm 0.08)$ V. Writing $(12.4700\pm 0.0824)$ V would be wrong by excess of precision: the four digits of the display are not four correct digits.

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d) At $x=0.25$ m: $E_{1}=\frac{(9.0\times 10^{9})(4.0\times 10^{-6})}{(0.25)^{2}}=5.76\times 10^{5}$ N/C toward $+x$; $E_{2}=\frac{(9.0\times 10^{9})(9.0\times 10^{-6})}{(0.25)^{2}}=1.296\times 10^{6}$ N/C toward $-x$. Net field: $7.2\times 10^{5}$ N/C toward $-x$ (toward $q_{1}$). Force on $q_{0}<0$: $F=|q_{0}|E=(5.0\times 10^{-6})(7.2\times 10^{5})=3.6$ N, directed OPPOSITE to the field, so toward $+x$, that is, toward $q_{2}$. The negative sign of the charge reverses the direction: forgetting this reversal is the classic trap.

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a) Cut the sphere into shells of thickness $dr'$, of volume $4\pi r'^{2}dr'$, each of constant density: $Q=\displaystyle\int_{0}^{R}\rho_{0}\dfrac{r'}{R}4\pi r'^{2}\,dr'=\dfrac{4\pi\rho_{0}}{R}\displaystyle\int_{0}^{R}r'^{3}dr'=\dfrac{4\pi\rho_{0}}{R}\cdot\dfrac{R^{4}}{4}=\pi\rho_{0}R^{3}$. Numerically: $\pi\times 2.50\times 10^{-6}\times(0.120)^{3}=1.36\times 10^{-8}$ C, that is $13.6$ nC. Note that the result is $\pi\rho_{0}R^{3}$ and not $\dfrac{4}{3}\pi\rho_{0}R^{3}$: the density being zero at the centre and greatest at the edge, the volume-weighted average is $\dfrac{3}{4}\rho_{0}$.

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d) Connecting to ground means imposing $V=0$ on the shell. The inner surface cannot change: it is locked by Gauss's law applied in the metal as long as $q$ stays in place, so it keeps $-3.00$ nC. It is the OUTER surface that adjusts, draining its charge to the ground until the potential vanishes, which requires an outer charge of zero. The field at $r=10.0$ cm therefore becomes zero, and it is zero everywhere outside. Grounding has made the shell electrically invisible from the outside.

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a) Parallel first: $R_{23}=\frac{R_{2}R_{3}}{R_{2}+R_{3}}=\frac{(12)(4)}{16}=3.0$ Ω. Then the series: $R_{eq}=6+3=9.0$ Ω.

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a) In steady state the capacitor no longer lets any current through: $R_{1}$ and $R_{2}$ form a voltage divider carrying $\dfrac{10.0}{9.00\times 10^{3}}=1.11$ mA. $v_{C}=R_{2}I=\dfrac{R_{2}}{R_{1}+R_{2}}\mathcal{E}=3.33$ V, not $10.0$ V.

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a) The ion feels the electric force $qE$ and the magnetic force $qvB$, set up in opposition. It goes through undeflected if they cancel: $qE=qvB$, so $v=\frac{E}{B}=\frac{8.0\times 10^{4}}{0.40}=2.0\times 10^{5}$ m/s. The charge has cancelled: the selector sorts on speed alone, whatever the mass or the charge.

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a) $\varepsilon=-N A\dfrac{dB}{dt}$ with $\dfrac{dB}{dt}=0.240-0.120\,t$. Hence $\varepsilon(t)=-200\times 2.50\times 10^{-3}\times(0.240-0.120t)=-0.120+0.0600\,t$, in volts. At $t=0$: $\varepsilon=-0.120$ V, of magnitude $120$ mV. At $t=2.00$ s: $\varepsilon=0$ exactly. At $t=4.00$ s: $\varepsilon=+0.120$ V, the same magnitude as at the start but the OPPOSITE sign. Notice that $NA=0.500$ m$^{2}$ here, which simplifies every calculation: the emf is simply half of $\dfrac{dB}{dt}$.

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a) Inside side (length $\ell$, parallel to the axis): $\vec{B}$ uniform and parallel, contribution $B\ell$. Outside side: the field is practically zero around a long solenoid (the distant turns compensate the near ones), zero contribution. Two perpendicular sides: $\vec{B}\perp d\vec{l}$ inside and $B\approx 0$ outside, zero contributions. Total: $\oint\vec{B}\cdot d\vec{l}=B\ell$. Enclosed current: the rectangle embraces $n\ell$ turns each carrying $I$, that is $I_{\text{enc}}=n\ell I$. Ampère: $B\ell=\mu_{0}n\ell I$, so $B=\mu_{0}nI$. Notice everything that has disappeared: $\ell$ (the field is uniform), the position of the inside side, and even the RADIUS of the solenoid.

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a) Any increase in $i$ increases the flux of the coil's own field through itself: by Lenz's law, the induced emf $e=-L\frac{di}{dt}$ OPPOSES this increase. A sudden jump in current would require an infinite emf: the current therefore builds up gradually. At $t=0^{+}$: $i=0$, the inductor behaves like an OPEN SWITCH (the whole voltage is across it). In steady state: $\frac{di}{dt}=0$, no emf any more, and the ideal inductor is a plain WIRE: $I_{\infty}=\frac{E}{R}=\frac{12}{6.0}=2.0$ A. This is the exact mirror of the capacitor, which acts as a wire at the start and an open circuit at the end.

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