Corrected exercises: magnetism and electromagnetic induction (203-SN2-RE)

$v$ and $B$ are perpendicular: $F=qvB=(1.6\times 10^{-19})(2.0\times 10^{6})(0.50)=1.6\times 10^{-13}$ N. Right hand: fingers toward the east ($v$), curled toward the north ($B$), the thumb points UP: the force is vertical and upward.

Weight: $mg=(1.67\times 10^{-27})(9.8)\approx 1.6\times 10^{-26}$ N. Ratio: $\frac{F}{mg}\approx 10^{13}$, ten trillion times the weight. Gravity is completely negligible in the dynamics of charged particles, as in electrostatics.

Same magnitude (same $|q|$, same velocity, same field), but the OPPOSITE direction: the negative charge reverses the sense, and the force on the electron points vertically down.

$F=qvB\sin 30^{\circ}=(1.6\times 10^{-13})(0.5)=8.0\times 10^{-14}$ N. The force vanishes when the velocity is PARALLEL (or antiparallel) to the field: $\sin 0^{\circ}=0$; a particle moving along the field lines feels nothing. Finally, the magnetic force stays perpendicular to the velocity at every instant: its work $W=Fd\cos 90^{\circ}$ is always zero, so the kinetic energy, and with it the magnitude of the velocity, cannot change. The magnetic field is a steering wheel, not an accelerator.

The force $qvB$, perpendicular to the velocity, has a constant magnitude ($v$ and $B$ do not change). A force of constant magnitude that is always perpendicular to the velocity is exactly a centripetal force: it bends the path into a circle without changing the magnitude of the velocity. Since the force stays perpendicular to the displacement at every instant, the magnetic field never does WORK: the kinetic energy of the proton is constant, only the direction of the velocity turns.

Newton's second law with the centripetal acceleration: $qvB=\frac{mv^{2}}{r}$. Solving: $r=\frac{mv}{qB}$. The faster or heavier the particle, the larger the circle; the stronger the field, the tighter it is.

$r=\frac{(1.67\times 10^{-27})(2.0\times 10^{6})}{(1.6\times 10^{-19})(0.50)}\approx 4.2\times 10^{-2}$ m, or about 4.2 cm. Period: $T=\frac{2\pi r}{v}=\frac{2\pi m}{qB}=\frac{2\pi(1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.50)}\approx 1.3\times 10^{-7}$ s.

In $T=\frac{2\pi m}{qB}$ the speed has cancelled: a fast particle traces a large circle, a slow one a small circle, but both complete their lap in the SAME time. The cyclotron exploits this synchronization: an alternating voltage of fixed frequency can accelerate the particle on every half-turn, without ever falling out of step as the radius grows. For an electron: the mass is 1836 times smaller, so the radius is 1836 times smaller (about 23 μm) and the period 1836 times shorter, with the rotation in the opposite sense (negative charge).

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.

The electric force $qE$ does not depend on the speed, but the magnetic force $qvB$ grows with it: for an ion that is too fast, magnetism wins, and the ion is deflected toward the MAGNETIC force. Too slow, and the electric force wins.

In the second region: $r=\frac{mv}{qB_{2}}$, so $m=\frac{rqB_{2}}{v}=\frac{(0.050)(1.6\times 10^{-19})(0.50)}{2.0\times 10^{5}}=2.0\times 10^{-26}$ kg. In atomic units: $\frac{2.0\times 10^{-26}}{1.66\times 10^{-27}}\approx 12$ u. This is very probably carbon 12 (singly ionized).

The radius is proportional to the mass ($v$, $q$ and $B_{2}$ unchanged): $r_{14}=5.0\times\frac{14}{12}\approx 5.8$ cm. The two isotopes, chemically identical (same electrons), hit the detector at distinct positions about 1.7 cm apart: the spectrometer sorts by INERTIA what chemistry cannot sort. This is the principle of carbon 14 dating and of isotope analysis.

$F=BIL=(0.20)(8.0)(1.5)=2.4$ N: a very real force, and it is what makes electric motors turn.

$B=\frac{\mu_{0}I}{2\pi d}=\frac{(4\pi\times 10^{-7})(20)}{2\pi(0.10)}=4.0\times 10^{-5}$ T: comparable to the Earth's field. A simple strong household current creates, at arm's length, a field of the same order as that of the whole planet: the magnetism of currents is surprisingly accessible.

Wire 1 creates at the position of wire 2 the field $B_{1}=4.0\times 10^{-5}$ T (computed in b). Force on 1 m of wire 2: $\frac{F}{L}=B_{1}I_{2}=(4.0\times 10^{-5})(20)=8.0\times 10^{-4}$ N/m. Right-hand rule: the field of wire 1 encircles it; at wire 2, the force $I\vec{L}\times\vec{B}$ points TOWARD wire 1. By symmetry (and Newton's third law), wire 1 is pulled toward wire 2: currents in the same direction ATTRACT.

The force depends only on the currents and the geometry: $\frac{F}{L}=\frac{\mu_{0}I_{1}I_{2}}{2\pi d}$. The former definition of the ampere rested on it: one ampere is the current that, in two parallel wires one metre apart, produces $2\times 10^{-7}$ N per metre, a purely mechanical definition, measurable with a balance. If one current is reversed, the right-hand rule reverses the force: opposite currents REPEL, same magnitude, opposite direction.

Area of one turn: $A=(0.080)(0.120)=9.6\times 10^{-3}$ m². Magnetic moment: $\mu=NIA=(50)(2.0)(9.6\times 10^{-3})=0.96$ A·m². The torque $\tau=\mu B\sin\theta$ is greatest for $\theta=90°$, that is, when the plane of the coil CONTAINS the field (the normal is perpendicular to it), and zero for $\theta=0°$ or $180°$, when the normal is aligned with the field.

$\tau_{max}=\mu B=(0.96)(0.30)=0.288$ N·m. At $\theta=30°$: $\tau=(0.288)(0.500)=0.144$ N·m. The torque always tends to REDUCE $\theta$, that is, to align the magnetic moment with the field: the coil pivots like a compass needle looking for north.

$U=-\mu B\cos\theta$ is least at $\theta=0°$, with $U=-0.288$ J: this is the STABLE equilibrium, moment aligned with the field. It is greatest at $\theta=180°$, with $U=+0.288$ J: UNSTABLE equilibrium, moment antiparallel. Work to flip the coil: $W=\Delta U=0.288-(-0.288)=0.576$ J, that is, $2\mu B$.

The torque vanishes precisely at alignment, and above all it CHANGES SIGN if the coil goes past that position: beyond it, the torque pulls the coil back. Left to itself, the coil would oscillate around $\theta=0°$ and then come to rest, a quarter turn after the start. The brush commutator reverses the direction of the current at every half-turn, exactly as the coil passes through alignment: the magnetic moment flips, the torque keeps the same sense of rotation, and the motion becomes continuous. The DC motor is therefore, above all, a switch that flips at the right moment.

At the equilibrium of the needle, the magnetic torque balances the restoring torque of the spring: $NIAB\sin\theta=k\alpha$. If we arrange for $\sin\theta=1$ at all times, what remains is $\alpha=\frac{NAB}{k}I$: the deflection is strictly proportional to the current, and the dial can be graduated linearly. But $\sin\theta$ stays equal to 1 only if the field remains parallel to the plane of the coil whatever its rotation: this is exactly what pole pieces hollowed into arcs of a circle produce, making the field RADIAL in the air gap. Without them, the graduation would go as $\sin\theta$ and the scale would be squeezed at large deflections.

The carriers that make up the current move at the drift speed $v_{d}$; the magnetic field exerts on them a force $qv_{d}B$ perpendicular to both the current and the field, hence directed across the strip. They pile up on one edge, leaving the other edge with the opposite charge. This separation creates a transverse electric field $E_{H}$ that pushes back on the next carriers: the build-up stops when the two forces balance, $qE_{H}=qv_{d}B$, that is, $E_{H}=v_{d}B$. The voltage between the edges is then $V_{H}=E_{H}w=v_{d}Bw$. Since $I=nqv_{d}A=nqv_{d}(wt)$, we get $v_{d}=\frac{I}{nqwt}$ and therefore $V_{H}=\frac{IB}{nqt}$: the width drops out.

$v_{d}=\frac{I}{nq\,wt}=\frac{10}{(8.5\times 10^{28})(1.6\times 10^{-19})(0.020)(1.0\times 10^{-4})}\approx 3.68\times 10^{-4}$ m/s. Voltage: $V_{H}=\frac{IB}{nqt}=\frac{(10)(1.2)}{(8.5\times 10^{28})(1.6\times 10^{-19})(1.0\times 10^{-4})}\approx 8.8\times 10^{-6}$ V, nine microvolts for ten amperes and more than a tesla. That is tiny, at the edge of the noise of an ordinary setup: in a metal, the density of carriers is so huge that each one moves very slowly, and the Hall voltage with it.

Take the current to the right and the field into the page. POSITIVE charges would go to the right and be pushed, say, upward: the top edge becomes positive. ELECTRONS produce the same current by moving to the left; the velocity reverses AND the charge changes sign, so the force $q\vec{v}\times\vec{B}$ keeps exactly the same direction: the electrons are pushed upward too. But this time the top edge becomes NEGATIVE. The two hypotheses therefore give Hall voltages of OPPOSITE SIGNS: the measured polarity directly reveals the sign of the carriers, which no resistance measurement can do. Historically, this is how it was confirmed that the current in common metals is carried by negative charges, and how it was discovered that some metals, such as zinc or beryllium, give a POSITIVE Hall coefficient: their carriers behave like positive charges, the holes, which only the quantum physics of bands explains.

$V_{H}=\frac{(0.010)(1.2)}{(1.0\times 10^{22})(1.6\times 10^{-19})(1.0\times 10^{-4})}=\frac{0.012}{0.16}=7.5\times 10^{-2}$ V, or 75 mV: a comfortable signal, readable without an amplifier. Copper, under the same 10 mA, would give $8.8\times 10^{-9}$ V, eight million times less. It all comes down to the $n$ in the denominator: a semiconductor has millions of times fewer carriers, hence much faster ones at the same current, hence a usable Hall voltage. That is why all Hall probes, the position sensors of brushless motors and crankshaft sensors are made of semiconductor.

$B=\frac{V_{H}nqt}{I}=\frac{(0.045)(1.0\times 10^{22})(1.6\times 10^{-19})(1.0\times 10^{-4})}{0.010}=\frac{(0.045)(0.16)}{0.010}=0.72$ T. In $V_{H}=\frac{IB}{nqt}$, neither the length nor the resistance appears: the length only sets the LONGITUDINAL voltage that must be applied to get the desired current, whereas the Hall voltage is transverse and is set only by the balance between the magnetic force and the electric force. As long as the current is imposed, the probe gives the same $V_{H}$ whether it is short or long, hot or cold: that is what makes it a reliable measuring instrument.

$v_{\parallel}=v\cos 30°=(4.0\times 10^{6})(0.866)\approx 3.46\times 10^{6}$ m/s and $v_{\perp}=v\sin 30°=2.0\times 10^{6}$ m/s. The magnetic force is $qvB\sin\theta$ and acts only on the PERPENDICULAR component: the component parallel to the field produces no force and therefore stays constant. The motion is the superposition of a uniform circle in the perpendicular plane and a uniform translation along the field: a helix, like the thread of a screw.

$r=\frac{mv_{\perp}}{qB}=\frac{(1.67\times 10^{-27})(2.0\times 10^{6})}{(1.6\times 10^{-19})(0.20)}\approx 0.104$ m. Period: $T=\frac{2\pi m}{qB}=\frac{2\pi(1.67\times 10^{-27})}{(1.6\times 10^{-19})(0.20)}\approx 3.28\times 10^{-7}$ s, independent of the angle of entry as of the speed. Pitch of the helix: $p=v_{\parallel}T=(3.46\times 10^{6})(3.28\times 10^{-7})\approx 1.14$ m.

Where the field lines tighten, they are no longer strictly parallel: they converge, and the force $q\vec{v}\times\vec{B}$ acquires a component directed TOWARD THE WEAK-FIELD REGION, which slows the motion along the axis. Moreover, since the magnetic force does no work, the total speed is fixed: as the field increases, the perpendicular part of the motion grows and the parallel part is used up, until it vanishes and then changes sign. The particle bounces back: this is the magnetic mirror. The Earth's field, stronger near the poles, forms precisely two mirrors facing each other: the solar-wind particles captured there shuttle from one hemisphere to the other, and these are the Van Allen belts. Those that dive low enough to strike the upper atmosphere excite oxygen and nitrogen there, which radiate: the polar auroras.

Cyclotron frequency: $f=\frac{1}{T}=\frac{qB}{2\pi m}=\frac{(1.6\times 10^{-19})(1.5)}{2\pi(1.67\times 10^{-27})}\approx 2.29\times 10^{7}$ Hz, or about 23 MHz. Exit speed, reached at the maximum radius: $v=\frac{qBR}{m}=\frac{(1.6\times 10^{-19})(1.5)(0.50)}{1.67\times 10^{-27}}\approx 7.19\times 10^{7}$ m/s. Energy: $K=\frac{1}{2}mv^{2}\approx 4.31\times 10^{-12}$ J, or $\frac{4.31\times 10^{-12}}{1.6\times 10^{-13}}\approx 27$ MeV.

$\frac{v}{c}=\frac{7.19\times 10^{7}}{3.0\times 10^{8}}\approx 0.24$: the proton is already moving at nearly a quarter of the speed of light, and its effective mass has increased by a few percent. Now the whole magic of the cyclotron rests on a CONSTANT period $T=\frac{2\pi m}{qB}$, which allows an accelerating voltage of fixed frequency. As soon as the relativistic mass grows, the period lengthens, the particle arrives late at the next gap and ends up being slowed instead of accelerated: the classical cyclotron tops out around a few tens of MeV for protons. The synchrocyclotron gradually lowers the accelerating frequency during the cycle to stay synchronized, at the price of a beam that is no longer continuous but pulsed. As for electrons, their mass is 1836 times smaller: they become relativistic at a few hundred keV, that is, practically from the start, and the betatron or the linear accelerator is preferred for them.

$\varepsilon=BLv=(0.50)(0.40)(5.0)=1.0$ V. Physically, every free charge in the rod, carried along at speed $v$ in the field $B$, feels a magnetic force $qvB$ along the rod: this force separates the charges and plays the role of a generator. Equivalently, the area of the circuit grows, so the flux $\Phi=BA$ grows, and Faraday gives $\varepsilon=\frac{\Delta\Phi}{\Delta t}=BLv$.

$I=\frac{\varepsilon}{R}=\frac{1.0}{0.80}=1.25$ A. Lenz's law: the induced current opposes the CAUSE that creates it, that is, the motion of the rod. The current, flowing through the rod in the field, produces a Laplace force directed AGAINST the velocity: a magnetic brake.

For a constant speed, the operator must exactly cancel this braking: $F=BIL=(0.50)(1.25)(0.40)=0.25$ N.

Mechanical: $P=Fv=(0.25)(5.0)=1.25$ W. Electrical: $P=\varepsilon I=(1.0)(1.25)=1.25$ W (or $RI^{2}=(0.80)(1.5625)=1.25$ W). Exact equality: every joule supplied by the muscle comes out as heat in the resistor. The rod on rails is an alternator reduced to its essentials: a mechanical-to-electrical converter, lossless here since nothing else rubs.

With no operator, the braking force $BIL$ persists as long as the rod moves: the rod decelerates, the emf and the current decrease with $v$, so the braking weakens too: the speed decays exponentially, without ever reversing. Lenz's law guarantees this sign: if the induced current PUSHED the rod instead of braking it, it would speed up by itself, creating energy out of nothing. Lenz's law is conservation of energy disguised as a sign rule.

$\Delta\Phi=\Delta B\times A=(0.60)(50\times 10^{-4})=3.0\times 10^{-3}$ Wb per turn.

Faraday, with the 200 turns in series: $\varepsilon=N\frac{\Delta\Phi}{\Delta t}=200\times\frac{3.0\times 10^{-3}}{0.30}=2.0$ V. Current: $I=\frac{\varepsilon}{R}=\frac{2.0}{8.0}=0.25$ A.

$Q=I\,\Delta t=(0.25)(0.30)=0.075$ C. In general: $Q=I\,\Delta t=\frac{\varepsilon}{R}\Delta t=\frac{N\Delta\Phi}{R\,\Delta t}\Delta t=\frac{N\Delta\Phi}{R}$: the duration has cancelled. Check: $\frac{(200)(3.0\times 10^{-3})}{8.0}=0.075$ C. The surprise: whether the field rises in 0.3 s or in 3 minutes, the SAME charge flows; only the current (the flow rate) changes. This is the principle of the ballistic fluxmeter: measuring a charge to learn a change in flux, with no stopwatch.

The flux grows toward the front of the coil: by Lenz's law, the induced current creates its own field OPPOSED to this growth, so it flows in the sense that, seen from the side toward which the field points, is clockwise (right-hand rule applied backward on the normal). Turning the coil a quarter turn in a fixed field also cancels the flux ($\cos 90^{\circ}=0$): same $\Delta\Phi$, same physics: an emf appears DURING the rotation. Repeating this rotation continuously is exactly an alternator: all the electricity of the grid comes from coils turned in fields.

$\Phi=BA\cos\omega t$ for one turn, so $N\Phi=NBA\cos\omega t$. Faraday: $\varepsilon=-\frac{d(N\Phi)}{dt}=NBA\omega\sin\omega t$: the emf is sinusoidal, greatest when the flux is ZERO (coil in the plane of the field, where the flux changes fastest) and zero when the flux is greatest. Angular speed: $\omega=\frac{2\pi(3600)}{60}\approx 377$ rad/s, so $f=60$ Hz, the frequency of the North American grid. Amplitude: $\varepsilon_{max}=NBA\omega=(100)(0.25)(0.0200)(377)\approx 1.88\times 10^{2}$ V.

$\varepsilon_{rms}=\frac{\varepsilon_{max}}{\sqrt{2}}=\frac{188.5}{1.414}\approx 1.33\times 10^{2}$ V. The rms value is the one that, in direct current, would dissipate the SAME average power in a resistor: it is therefore the only one that allows AC and DC to be compared, and it is the one that is displayed. An outlet at 120 V rms really does oscillate between $+170$ V and $-170$ V, which explains why the insulation must withstand the peak value, not the displayed value.

$I_{max}=\frac{\varepsilon_{max}}{R}=\frac{188.5}{50}\approx 3.77$ A. Average power: $P=\frac{\varepsilon_{rms}^{2}}{R}=\frac{\varepsilon_{max}^{2}}{2R}=\frac{35531}{100}\approx 3.55\times 10^{2}$ W (the factor 2 comes from the average of $\sin^{2}$, which is $\frac{1}{2}$). Average torque: the turbine supplies all this power mechanically, $P=\tau\omega$, so $\tau=\frac{355.3}{377}\approx 0.94$ N·m. The more current you draw, the harder the turbine has to work: this is physically why a power plant burns more steam when demand rises.

As it falls, the magnet changes the flux through each horizontal slice of the tube. Faraday induces circular currents there, the eddy currents, and Lenz fixes their direction: they create a field that opposes the CHANGE, so they push the magnet away from below and hold it back from above. The result is a braking force that grows with speed until it balances the weight: the magnet then descends at terminal speed, slowly. Copper does not need to be magnetic: it only needs to be a good CONDUCTOR for the induced currents to be strong; in plastic, an insulator, no current can flow and the fall is free. Same mechanism in a transformer: a solid core would host eddy currents that would heat it for nothing. It is laminated into sheets insulated from one another, perpendicular to the path of these currents, which breaks them up and drastically reduces the losses.

The braking comes from closed LOOPS of current in the disc. The radial slots cut these loops: the currents can no longer go all the way round, their intensity collapses, and so does the braking force. The disc keeps turning almost freely, which incidentally proves that it was the circulation of current, and not the magnetization of the metal, that braked it. Conversely, an induction cooktop seeks precisely to maximize these currents: a coil carrying an alternating current of a few tens of kilohertz induces very strong eddy currents in the ferromagnetic bottom of the pot, which heat it by the Joule effect. The heat is born INSIDE the pot, not under it: it is the same phenomenon, once treated as a loss, once exploited as a heater.