Corrected practice final exam: electricity and magnetism, extended course (203-SN2-RE, formerly 203-NYB)

Between the two charges, the field of $q_{1}$ points to the right (it is positive, so it repels) and that of $q_{2}$ also points to the right (it is negative, so it attracts): the two contributions add and cannot cancel. To the left of $q_{1}$, we are closer to the charge of larger magnitude, so its field necessarily dominates there. The point we are looking for is therefore to the right of $q_{2}$, on the side of the weaker charge.

Let $x>0.30$. Equal magnitudes are written $\dfrac{k\left|q_{1}\right|}{x^{2}}=\dfrac{k\left|q_{2}\right|}{(x-0.30)^{2}}$. Since $\left|q_{1}\right|=4\left|q_{2}\right|$, we get $\dfrac{4}{x^{2}}=\dfrac{1}{(x-0.30)^{2}}$, that is $\dfrac{2}{x}=\dfrac{1}{x-0.30}$ after taking positive roots. So $2(x-0.30)=x$, hence $x=0.60\ \mathrm{m}$. Check: both magnitudes equal $99.9\ \mathrm{V/m}$.

$V=k\dfrac{q_{1}}{0.15}+k\dfrac{q_{2}}{0.15}=\dfrac{k\left(4.0-1.0\right)\times 10^{-9}}{0.15}=\dfrac{8.99\times 10^{9}\times 3.0\times 10^{-9}}{0.15}=179.8\ \mathrm{V}$.

We look for $x$ between 0 and $0.30$ such that $\dfrac{4}{x}=\dfrac{1}{0.30-x}$, that is $4(0.30-x)=x$, so $1.20=5x$ and $x=0.24\ \mathrm{m}$. The potential is zero there because the two contributions, one positive and one negative, cancel: they are scalars and add algebraically. The field, on the other hand, is a vector: at this point both contributions point to the right, so they add instead of cancelling. A zero potential never implies a zero field, and the converse is just as false.

$U=\dfrac{kq_{1}q_{2}}{r}=\dfrac{8.99\times 10^{9}\times 4.0\times 10^{-9}\times\left(-1.0\times 10^{-9}\right)}{0.30}=-1.20\times 10^{-7}\ \mathrm{J}$. The negative sign means that the system is bound: it would take $1.20\times 10^{-7}\ \mathrm{J}$ to pull the two charges apart to infinity. This is consistent with the fact that charges of opposite signs attract.

Gauss's law gives $\Phi=\dfrac{Q_{\mathrm{int}}}{\varepsilon_{0}}=\dfrac{6.0\times 10^{-6}}{8.854\times 10^{-12}}=6.78\times 10^{5}\ \mathrm{N\cdot m^{2}/C}$. This flux does not depend on the radius chosen as long as it exceeds $R$: the enclosed charge is then always $Q$. This is exactly what the law asserts, and it is what makes it so powerful.

By spherical symmetry, $E$ is radial and has constant magnitude on the Gaussian surface, so $\Phi=E\times 4\pi r^{2}$, hence $E=\dfrac{Q}{4\pi\varepsilon_{0}r^{2}}=\dfrac{kQ}{r^{2}}=\dfrac{8.99\times 10^{9}\times 6.0\times 10^{-6}}{0.20^{2}}=1.35\times 10^{6}\ \mathrm{V/m}$. The graph does give about $1.35\ \mathrm{MV/m}$ at $r=0.20$. Outside, the sphere behaves exactly like a point charge placed at its centre.

For $r<R$, the enclosed charge is the corresponding fraction of the volume: $Q_{\mathrm{int}}=Q\dfrac{\frac{4}{3}\pi r^{3}}{\frac{4}{3}\pi R^{3}}=Q\dfrac{r^{3}}{R^{3}}$. Gauss's law then gives $E\times 4\pi r^{2}=\dfrac{Q r^{3}}{\varepsilon_{0}R^{3}}$, so $E=\dfrac{Qr}{4\pi\varepsilon_{0}R^{3}}=\dfrac{kQr}{R^{3}}$.

$E=\dfrac{8.99\times 10^{9}\times 6.0\times 10^{-6}\times 0.050}{0.10^{3}}=2.70\times 10^{6}\ \mathrm{V/m}$. At the surface, $E=\dfrac{kQ}{R^{2}}=5.39\times 10^{6}\ \mathrm{V/m}$. The ratio is exactly 2, as expected: inside, $E$ is proportional to $r$, and $0.050$ is half of $R$.

The field is maximal exactly at the surface, at $r=R$, where it equals $5.39\times 10^{6}\ \mathrm{V/m}$. Continuity checks out there: the inside law gives $\dfrac{kQR}{R^{3}}=\dfrac{kQ}{R^{2}}$ and the outside law gives $\dfrac{kQ}{R^{2}}$, the same value. The field grows linearly up to the surface and then falls off as $\dfrac{1}{r^{2}}$, which explains the shape of the graph: a straight line followed by a decreasing hyperbola, joined without a jump.

Cut the ring into small charge elements. Each element has a diametrically opposite partner, which carries the same charge and lies at the same distance from the point on the axis. The components of their fields perpendicular to the axis are opposite and cancel in pairs, while their axial components add. The resultant field is therefore directed along the axis.

$kQ=8.99\times 10^{9}\times 8.0\times 10^{-9}=71.92$. At $z=0.10$: $z^{2}+a^{2}=0.020$ and $\left(0.020\right)^{3/2}=2.828\times 10^{-3}$, so $E=\dfrac{71.92\times 0.10}{2.828\times 10^{-3}}=2.54\times 10^{3}\ \mathrm{V/m}$. For the potential, $\sqrt{0.020}=0.1414$, so $V=\dfrac{71.92}{0.1414}=509\ \mathrm{V}$.

At $z=0$, the formula gives $E=0$: at the centre, each charge element is balanced by its opposite, so the field is zero. The potential there, however, is $V=\dfrac{kQ}{a}=\dfrac{71.92}{0.10}=719\ \mathrm{V}$, its maximum value. This is a clear example of the fact that a zero field does not imply a zero potential: the potential is a scalar, all the contributions are positive here and add up, whereas the field vectors cancel.

Differentiate $E$. With $u(z)=z$ and $v(z)=\left(z^{2}+a^{2}\right)^{3/2}$, we have $v'(z)=3z\left(z^{2}+a^{2}\right)^{1/2}$, so $E'=kQ\dfrac{\left(z^{2}+a^{2}\right)^{3/2}-3z^{2}\left(z^{2}+a^{2}\right)^{1/2}}{\left(z^{2}+a^{2}\right)^{3}}$. Factoring $\left(z^{2}+a^{2}\right)^{1/2}$ out of the numerator: $E'=kQ\dfrac{z^{2}+a^{2}-3z^{2}}{\left(z^{2}+a^{2}\right)^{5/2}}=kQ\dfrac{a^{2}-2z^{2}}{\left(z^{2}+a^{2}\right)^{5/2}}$. The denominator is positive, so $E'$ vanishes when $a^{2}=2z^{2}$, that is $z=\dfrac{a}{\sqrt{2}}=0.0707\ \mathrm{m}$, changing sign from plus to minus: this is indeed a maximum. Then $z^{2}+a^{2}=\dfrac{3a^{2}}{2}=0.015$, whose power $3/2$ is $1.837\times 10^{-3}$, and $E_{\max}=\dfrac{71.92\times 0.0707}{1.837\times 10^{-3}}=2.77\times 10^{3}\ \mathrm{V/m}$.

When $z$ becomes large compared with $a$, we can neglect $a^{2}$ next to $z^{2}$: $\left(z^{2}+a^{2}\right)^{3/2}\approx\left(z^{2}\right)^{3/2}=z^{3}$, so $E\approx\dfrac{kQz}{z^{3}}=\dfrac{kQ}{z^{2}}$, which is exactly the field of a point charge $Q$. Check at $z=1.0\ \mathrm{m}$: the exact formula gives $70.9\ \mathrm{V/m}$ and the point-charge approximation $71.9\ \mathrm{V/m}$, a difference of only 1.5 percent, even though $z$ is just ten times $a$.

$R_{23}=\dfrac{R_{2}R_{3}}{R_{2}+R_{3}}=\dfrac{6.0\times 3.0}{9.0}=2.0\ \Omega$. The total resistance is $R_{\mathrm{total}}=r+R_{1}+R_{23}=0.50+4.0+2.0=6.5\ \Omega$.

$I=\dfrac{\mathcal{E}}{R_{\mathrm{total}}}=\dfrac{12.0}{6.5}=1.846\ \mathrm{A}$. The voltage across the battery terminals is $U=\mathcal{E}-rI=12.0-0.50\times 1.846=11.08\ \mathrm{V}$: it is lower than the emf, precisely because of the internal resistance.

The voltage across the parallel combination is $U_{23}=R_{23}I=2.0\times 1.846=3.692\ \mathrm{V}$. It is the same for both branches, so $I_{2}=\dfrac{3.692}{6.0}=0.615\ \mathrm{A}$ and $I_{3}=\dfrac{3.692}{3.0}=1.231\ \mathrm{A}$. Their sum is $1.846\ \mathrm{A}$, equal to the total current: the junction rule checks out. Note that the smaller resistance carries the larger current, in the ratio $\dfrac{I_{3}}{I_{2}}=\dfrac{R_{2}}{R_{3}}=2$: the currents are inversely proportional to the resistances.

$P_{\mathcal{E}}=\mathcal{E}I=12.0\times 1.846=22.15\ \mathrm{W}$. Next, $P_{r}=rI^{2}=0.50\times 1.846^{2}=1.70\ \mathrm{W}$, $P_{1}=R_{1}I^{2}=4.0\times 3.408=13.63\ \mathrm{W}$ and $P_{23}=R_{23}I^{2}=2.0\times 3.408=6.82\ \mathrm{W}$. The three add up to $1.70+13.63+6.82=22.15\ \mathrm{W}$, equal to the power supplied: the balance is closed.

The fraction lost in the battery is $\dfrac{1.70}{22.15}=7.7$ percent. To reduce it, the current must be reduced, so the external resistance must be increased, or a battery of lower internal resistance chosen. Note that this loss grows as $I^{2}$ while the power supplied grows only as $I$: the more current is drawn, the larger the proportion wasted in the battery.

The bridge is balanced when $\dfrac{R_{1}}{R_{2}}=\dfrac{R_{3}}{R_{4}}$, which makes the voltage between $A$ and $B$ zero. Here $\dfrac{100}{200}=0.500$ while $\dfrac{300}{400}=0.750$: the condition is not satisfied, the bridge is unbalanced.

Each branch is a simple voltage divider. $V_{A}=10.0\times\dfrac{R_{2}}{R_{1}+R_{2}}=10.0\times\dfrac{200}{300}=6.667\ \mathrm{V}$ and $V_{B}=10.0\times\dfrac{R_{4}}{R_{3}+R_{4}}=10.0\times\dfrac{400}{700}=5.714\ \mathrm{V}$. So $U_{AB}=V_{A}-V_{B}=0.952\ \mathrm{V}$.

With the source short-circuited, the point $A$ is tied to ground through $R_{1}$ and $R_{2}$ in parallel, and the point $B$ through $R_{3}$ and $R_{4}$ in parallel. These two combinations are in series between $A$ and $B$, so $R_{\mathrm{eq}}=\dfrac{100\times 200}{300}+\dfrac{300\times 400}{700}=66.7+171.4=238.1\ \Omega$.

The circuit seen by the galvanometer is a voltage source with open-circuit voltage $0.952\ \mathrm{V}$ and internal resistance $238.1\ \Omega$. So $I_{g}=\dfrac{0.952}{238.1+50}=\dfrac{0.952}{288.1}=3.31\times 10^{-3}\ \mathrm{A}$, or about $3.31\ \mathrm{mA}$.

Balance requires $\dfrac{R_{1}}{R_{2}}=\dfrac{R_{3}}{R_{4}}$, so $R_{4}=\dfrac{R_{2}R_{3}}{R_{1}}=\dfrac{200\times 300}{100}=600\ \Omega$. The method is precise because at balance we do not measure a value, we observe a cancellation: the result then depends neither on the source voltage, nor on the resistance of the galvanometer, nor on its calibration. It depends only on a ratio of known resistances, and detecting a zero can be done with far greater sensitivity than reading a deflection.

The cross-sectional area is $A=\pi\left(\dfrac{d}{2}\right)^{2}=\pi\left(1.0\times 10^{-3}\right)^{2}=3.14\times 10^{-6}\ \mathrm{m^{2}}$. The length to use is the one travelled by the current, so the trip out AND back: $L=2\times 150=300\ \mathrm{m}$. Hence $R=\dfrac{\rho L}{A}=\dfrac{1.68\times 10^{-8}\times 300}{3.14\times 10^{-6}}=1.60\ \Omega$. Forgetting the return is the classic mistake and halves the result.

$I=\dfrac{P}{U}=\dfrac{1500}{120}=12.5\ \mathrm{A}$. The power lost is $P_{\mathrm{loss}}=RI^{2}=1.60\times 12.5^{2}=251\ \mathrm{W}$, which is $16.7$ percent of the useful power. The voltage drop is $RI=1.60\times 12.5=20.1\ \mathrm{V}$, which is considerable: the workshop actually receives only about $100\ \mathrm{V}$.

At $240\ \mathrm{V}$, the current falls to $I=\dfrac{1500}{240}=6.25\ \mathrm{A}$, half as much. The loss becomes $P_{\mathrm{loss}}=1.60\times 6.25^{2}=62.7\ \mathrm{W}$, a quarter of the previous value and only $4.2$ percent of the useful power. The gain comes from the loss varying as the square of the current: doubling the voltage halves the current and so divides the loss by four. This is the whole reason for high-voltage transmission lines.

We need $RI^{2}<50$ with $I=12.5\ \mathrm{A}$, so $R<\dfrac{50}{156.25}=0.320\ \Omega$. Since $R=\dfrac{\rho L}{A}$, we get $A>\dfrac{1.68\times 10^{-8}\times 300}{0.320}=1.575\times 10^{-5}\ \mathrm{m^{2}}$. As $A=\pi\dfrac{d^{2}}{4}$, it follows that $d>2\sqrt{\dfrac{A}{\pi}}=2\sqrt{5.01\times 10^{-6}}=4.48\times 10^{-3}\ \mathrm{m}$, or about $4.5\ \mathrm{mm}$. The diameter must therefore be more than doubled, because the area, and hence the resistance, depends on the square of the diameter.

$R_{60}=R_{20}\left[1+\alpha\left(T-T_{0}\right)\right]=1.60\times\left(1+3.9\times 10^{-3}\times 40\right)=1.60\times 1.156=1.85\ \Omega$, which is $15.6$ percent more. The runaway mechanism is as follows: a larger resistance dissipates more power at constant current, which heats the conductor further, which raises its resistance again. In a poorly ventilated or undersized cable, this positive feedback loop can lead to degradation of the insulation, and this is precisely why electrical codes impose a minimum gauge for each current rating.

The work of the electric force supplies the kinetic energy: $K=qU=1.602\times 10^{-19}\times 2000=3.204\times 10^{-16}\ \mathrm{J}$, which is exactly $2000\ \mathrm{eV}$ by the definition of the electronvolt. From $K=\dfrac{1}{2}mv^{2}$ we get $v=\sqrt{\dfrac{2qU}{m}}=\sqrt{\dfrac{2\times 3.204\times 10^{-16}}{1.673\times 10^{-27}}}=6.19\times 10^{5}\ \mathrm{m/s}$.

The magnetic force $\vec{F}=q\vec{v}\times\vec{B}$ is perpendicular to the velocity, so it does not change the magnitude of $\vec{v}$ but only its direction: it is a centripetal force of constant magnitude $qvB$. A motion at constant speed under a constant centripetal force is uniform circular motion. Writing $qvB=\dfrac{mv^{2}}{R}$ gives $R=\dfrac{mv}{qB}=\dfrac{1.673\times 10^{-27}\times 6.19\times 10^{5}}{1.602\times 10^{-19}\times 0.50}=1.29\times 10^{-2}\ \mathrm{m}$, or about $1.3\ \mathrm{cm}$.

$T=\dfrac{2\pi R}{v}=\dfrac{2\pi m}{qB}=\dfrac{2\pi\times 1.673\times 10^{-27}}{1.602\times 10^{-19}\times 0.50}=1.31\times 10^{-7}\ \mathrm{s}$, and $f=\dfrac{1}{T}=7.62\times 10^{6}\ \mathrm{Hz}$. The expression $\dfrac{2\pi m}{qB}$ contains neither $v$ nor $R$: the period depends only on the particle and the field. A faster proton traces a larger circle, but in exactly the same time. This property is what makes the cyclotron possible, since an alternating voltage of fixed frequency stays in step with the particle.

The radius $R=\dfrac{mv}{qB}$ is inversely proportional to $B$: it is therefore halved, to $0.646\ \mathrm{cm}$. The period $T=\dfrac{2\pi m}{qB}$ is also halved, to $6.56\times 10^{-8}\ \mathrm{s}$. The speed does not change.

No. The magnetic force is at every instant perpendicular to the displacement, so its work is zero, and by the work-energy theorem the kinetic energy stays constant. A magnetic field can bend a path, but never accelerate a particle in the energy sense. This is why every accelerator needs an electric field to supply the energy, the magnetic field serving only to guide.

Only the part still in the field contributes to the flux. During the exit, the immersed length decreases at the speed $v$, so $\left|\dfrac{\mathrm{d}\Phi}{\mathrm{d}t}\right|=N B L v$. Thus $\mathcal{E}=50\times 0.40\times 0.10\times 3.0=6.0\ \mathrm{V}$ and $I=\dfrac{\mathcal{E}}{R}=\dfrac{6.0}{2.0}=3.0\ \mathrm{A}$.

The flux through the coil decreases, since the immersed area decreases. By Lenz's law, the induced current opposes this decrease: it flows in the direction that creates, inside the coil, a field in the same direction as the external field. With $\vec{B}$ into the page, as in the solution figure, the current therefore flows clockwise; if $\vec{B}$ came out of the page, it would flow counterclockwise. In both cases it acts so as to maintain the flux.

The side still in the field carries the current $I$ over the length $L$, for each of the $N$ turns: $F=N B I L=50\times 0.40\times 3.0\times 0.10=6.0\ \mathrm{N}$. This force opposes the motion, again by Lenz's law, so the person pulling must supply an equal and opposite force. The mechanical power is $P_{\mathrm{mech}}=Fv=6.0\times 3.0=18\ \mathrm{W}$.

$P_{\mathrm{elec}}=\mathcal{E}I=6.0\times 3.0=18\ \mathrm{W}$, or equivalently $RI^{2}=2.0\times 9.0=18\ \mathrm{W}$. This is exactly the mechanical power of part c). It is no coincidence: all the energy supplied by the person pulling is converted into heat in the resistance of the coil, since the speed is constant and there is therefore no change in kinetic energy. Lenz's law is here the expression of conservation of energy.

The coil is completely out once it has travelled its own width, so $\Delta t=\dfrac{L}{v}=\dfrac{0.10}{3.0}=3.33\times 10^{-2}\ \mathrm{s}$. The energy dissipated is $W=P\Delta t=18\times 3.33\times 10^{-2}=0.60\ \mathrm{J}$. Check using the work of the applied force: $W=F\times L=6.0\times 0.10=0.60\ \mathrm{J}$. The two calculations agree, which confirms the energy balance.