Corrected exercises: sources of the magnetic field and the Biot-Savart law (203-SN2-RE)

The element at $P$ creates at $M$ a field $d\vec{B}$ perpendicular to $PM$ and lying in the plane $(OPM)$: it has one component along the axis and one component perpendicular to the axis. The symmetric element at $P'$ (same distance $r$, opposite current) creates a field of the same magnitude whose perpendicular component is exactly OPPOSITE, while the axial components add. Going around the whole loop, every element finds its partner in this way: the perpendicular components cancel in pairs and only the axial component survives. With the current of the figure (⊙ at the top, ⊗ at the bottom), the right-hand rule gives a field directed from $O$ toward $M$.

$d\vec{l}$ is tangent to the loop at $P$: it is perpendicular to the radius $OP$ (a tangent is perpendicular to the radius) and perpendicular to the axis $OM$ (it lies in the plane of the loop, to which the axis is perpendicular). Being perpendicular to two intersecting lines of the plane $(OPM)$, it is perpendicular to that whole plane, hence in particular to $PM$. In the Biot-Savart law, $\sin\theta=\sin 90^{\circ}=1$: $dB=\frac{\mu_{0}}{4\pi}\frac{I\,dl}{r^{2}}$. And since all the points of the loop are at the SAME distance $r$ from $M$, this magnitude is the same for every element: that is what makes the summation immediate.

$d\vec{B}$ is perpendicular to $PM$ in the plane $(OPM)$; since $PM$ makes the angle $\alpha$ with the axis, $d\vec{B}$ makes the angle $90^{\circ}-\alpha$ with the axis and its axial component is $dB\sin\alpha$. The sum over the loop gives $B=\frac{\mu_{0}I\sin\alpha}{4\pi r^{2}}\times\sum dl=\frac{\mu_{0}I\sin\alpha}{4\pi r^{2}}\times 2\pi R=\frac{\mu_{0}IR\sin\alpha}{2r^{2}}$. The triangle $OPM$, right-angled at $O$, supplies the missing geometry: $\sin\alpha=\frac{R}{r}$, so $r=\frac{R}{\sin\alpha}$. Substituting: $B=\frac{\mu_{0}IR\sin\alpha\sin^{2}\alpha}{2R^{2}}=\frac{\mu_{0}I\sin^{3}\alpha}{2R}$. Only three quantities, as requested: $I$, $\alpha$ and $R$.

At the centre, $M$ is at $O$: $\alpha=90^{\circ}$, $\sin^{3}\alpha=1$, and we recover the known formula $B=\frac{\mu_{0}I}{2R}$. In terms of $x=OM$: $r^{2}=R^{2}+x^{2}$ (Pythagoras) and $\sin\alpha=\frac{R}{\sqrt{R^{2}+x^{2}}}$, hence $B=\frac{\mu_{0}IR^{2}}{2(R^{2}+x^{2})^{3/2}}$. For $x\gg R$, $(R^{2}+x^{2})^{3/2}\approx x^{3}$: $B\approx\frac{\mu_{0}IR^{2}}{2x^{3}}$, a $1/x^{3}$ decrease, much faster than the $1/d$ of the infinite wire. The reason: seen from far away, the loop is a magnetic DIPOLE, a tiny magnet whose two faces (north and south) almost cancel each other; the infinite wire never closes on itself and its influence reaches much farther.

$B=\frac{\mu_{0}I}{2R}=\frac{(4\pi\times 10^{-7})(12)}{2(0.080)}\approx 9.4\times 10^{-5}$ T: barely twice Earth's field. A single loop is a rather modest source, which is why we wind coils.

The 100 turns occupy practically the same position and carry the same current: each creates the same field at the centre, and the fields, all in the same direction, add up. $B=N\frac{\mu_{0}I}{2R}=100\times 9.4\times 10^{-5}\approx 9.4\times 10^{-3}$ T. Winding a coil multiplies the source without increasing the current.

The electron passes the same spot on every revolution, that is $\frac{v}{2\pi r}$ times per second: the charge crossing a cross-section per second is $I=\frac{ev}{2\pi r}=\frac{(1.6\times 10^{-19})(2.2\times 10^{6})}{2\pi(5.3\times 10^{-11})}\approx 1.1\times 10^{-3}$ A, a mere milliampere. Field at the centre of this atomic “loop”: $B=\frac{\mu_{0}I}{2r}=\frac{(4\pi\times 10^{-7})(1.06\times 10^{-3})}{2(5.3\times 10^{-11})}\approx 12.5$ T: stronger than most MRI magnets.

$I=\frac{2RB}{\mu_{0}}=\frac{2(0.080)(12.5)}{4\pi\times 10^{-7}}\approx 1.6\times 10^{6}$ A: 1.6 million amperes in our 8 cm loop! The key is in $B=\frac{\mu_{0}I}{2R}$: the field is inversely proportional to the RADIUS. The atomic radius is about a billion times smaller than the lab loop, and this proximity more than makes up for the weakness of the current. In magnetism, being close to the source beats being powerful.

The Biot-Savart law involves $dl\sin\theta$, where $\theta$ is the angle between the element $d\vec{l}$ and the line joining the element to the point under study. If the extension of the segment passes through that point, this line lies along the segment itself: $\theta=0$ or $180°$, so $\sin\theta=0$ and every element contributes nothing. Consequence for the figure: the two RADIAL segments, whose extensions pass through $O$, do not count. Only the two arcs create a field at $O$, which reduces the problem to two terms.

On an arc, every element $d\vec{l}$ is tangent to the circle, hence perpendicular to the radius joining it to the centre: $\sin\theta=1$ for all of them. Moreover, all the elements are at the same distance $R$ and their fields all have the same direction (perpendicular to the plane of the arc) and the same sense: the magnitudes simply add. $B=\frac{\mu_{0}I}{4\pi R^{2}}\sum dl=\frac{\mu_{0}I}{4\pi R^{2}}(R\theta)=\frac{\mu_{0}I\theta}{4\pi R}$. For a complete loop, $\theta=2\pi$: $B=\frac{\mu_{0}I(2\pi)}{4\pi R}=\frac{\mu_{0}I}{2R}$, the known formula. The field of an arc is therefore only the FRACTION $\frac{\theta}{2\pi}$ of the field of the whole loop.

Each arc spans $\theta=\frac{\pi}{2}$. Inner arc: $B_{a}=\frac{\mu_{0}I\theta}{4\pi a}=\frac{(10^{-7})(8.0)(1.571)}{0.030}\approx 4.19\times 10^{-5}$ T. Outer arc: $B_{b}=\frac{(10^{-7})(8.0)(1.571)}{0.060}\approx 2.09\times 10^{-5}$ T, half as much, since the field of an arc varies as $\frac{1}{R}$ at equal angle. The current runs through the two arcs in OPPOSITE senses (it circulates one way along one arc and the other way along the other): the two fields are opposite and we subtract. $B_{O}=4.19\times 10^{-5}-2.09\times 10^{-5}\approx 2.09\times 10^{-5}$ T, in the direction set by the INNER arc, the closer and therefore stronger one.

Point $P$ lies on the perpendicular bisector: the two ends of the wire are seen symmetrically, at the angles $\theta_{2}=+\theta_{max}$ and $\theta_{1}=-\theta_{max}$, so that $\sin\theta_{2}-\sin\theta_{1}=2\sin\theta_{max}$. The right triangle gives $\sin\theta_{max}=\frac{L}{\sqrt{L^{2}+d^{2}}}$ with $L=0.10$ m (the half-length) and $d=0.050$ m. Hence $B=\frac{\mu_{0}I}{4\pi d}\cdot\frac{2L}{\sqrt{L^{2}+d^{2}}}=\frac{\mu_{0}IL}{2\pi d\sqrt{L^{2}+d^{2}}}$. Numerically: $\sqrt{L^{2}+d^{2}}=\sqrt{0.0125}\approx 0.1118$ m, so $\sin\theta_{max}\approx 0.894$ and $B=\frac{(10^{-7})(10)}{0.050}\times 2(0.894)\approx 3.58\times 10^{-5}$ T.

Infinite wire at the same distance: $B_{\infty}=\frac{\mu_{0}I}{2\pi d}=\frac{(2\times 10^{-7})(10)}{0.050}=4.0\times 10^{-5}$ T. The ratio is exactly $\sin\theta_{max}\approx 0.894$: the finite wire gives 89 % of the infinite wire, an 11 % error for a wire that is nevertheless four times longer than the distance. To be within 1 %, we need $\sin\theta_{max}\geq 0.99$, that is $L^{2}\geq 0.9801(L^{2}+d^{2})$, so $L\geq d\sqrt{\frac{0.9801}{0.0199}}\approx 7.0\,d=0.35$ m: a wire 70 cm long observed from 5 cm. Remember the rule: the infinite-wire approximation is good only if the wire extends at least seven times the observation distance ON EACH SIDE.

At the centre of an isolated coil: $B=\frac{\mu_{0}NI}{2R}=\frac{(4\pi\times 10^{-7})(100)(2.0)}{2(0.10)}\approx 1.26\times 10^{-3}$ T.

$M$ is at $x=\frac{R}{2}$ from each coil. For one coil: $B=\frac{\mu_{0}NIR^{2}}{2(R^{2}+\frac{R^{2}}{4})^{3/2}}=\frac{\mu_{0}NIR^{2}}{2(\frac{5R^{2}}{4})^{3/2}}=\frac{\mu_{0}NIR^{2}}{2R^{3}(\frac{5}{4})^{3/2}}=\left(\frac{4}{5}\right)^{3/2}\frac{\mu_{0}NI}{2R}$. The two coils, carrying current in the same direction, create fields of the same direction at $M$: we double. $B_{M}=\left(\frac{4}{5}\right)^{3/2}\frac{\mu_{0}NI}{R}=0.716\times\frac{(4\pi\times 10^{-7})(100)(2.0)}{0.10}\approx 1.80\times 10^{-3}$ T. Numerically, each coil contributes $8.99\times 10^{-4}$ T.

Coil 1 ($x=0.025$ m): $B_{1}=\frac{\mu_{0}NIR^{2}}{2(0.010625)^{3/2}}\approx 1.15\times 10^{-3}$ T. Coil 2 ($x=0.075$ m): $B_{2}=\frac{\mu_{0}NIR^{2}}{2(0.015625)^{3/2}}\approx 6.43\times 10^{-4}$ T. Total: $B_{Q}\approx 1.79\times 10^{-3}$ T. Relative difference from $B_{M}$: about 0.4 %. Moving away from the midpoint by half of the available distance, the field has changed by only 0.4 %: what one coil loses, the other regains almost exactly. The separation $d=R$ is precisely the one that achieves this compensation best.

This is THE standard arrangement for an accessible uniform field: between the coils, an entire experiment (electron beam, probe, sample) can be placed in a known, constant field, and Earth's field can also be cancelled there. If one coil is reversed, the two fields at $M$ become opposite and of equal magnitude: $B_{M}=0$. The field is no longer uniform but a GRADIENT, zero at the centre and growing on either side: this is the anti-Helmholtz arrangement, used to trap cold atoms.

By symmetry, the field lines of an infinite wire are circles centred on the wire, and the magnitude of $B$ is the same at every point of a given circle (nothing distinguishes one point from another along the circle). On this loop, $\vec{B}$ is everywhere parallel to $d\vec{l}$ and of constant magnitude: $\oint\vec{B}\cdot d\vec{l}=B\times 2\pi d$. Ampère's law gives $B(2\pi d)=\mu_{0}I$, hence $B=\frac{\mu_{0}I}{2\pi d}$. Without this symmetry, $B$ would not come out of the integral and the law, which is always true, would become useless for calculation.

$B=\frac{\mu_{0}I}{2\pi d}=\frac{(2\times 10^{-7})(25)}{0.020}=2.5\times 10^{-4}$ T, five times Earth's field: a compass placed near a jumper cable literally loses its north.

Right hand: between the wires, the field of wire 1 and that of wire 2 point in OPPOSITE directions (perpendicular to the segment, one upward, the other downward): cancellation is possible there. Outside the segment, the two fields point in the same direction and can never cancel. Position: $\frac{\mu_{0}I_{1}}{2\pi x}=\frac{\mu_{0}I_{2}}{2\pi(0.20-x)}$ gives $6(0.20-x)=3x$, so $x=\frac{6}{9}(0.20)\approx 13.3$ cm from wire 1, that is 6.7 cm from wire 2: the point of zero field takes refuge near the WEAKER current, where the stronger one, weakened by distance, can be matched.

Opposite currents: between the wires, the two fields now point in the SAME direction and add, so no cancellation is possible. Outside they oppose each other; but on the side of wire 1 (the strong one), the field of the 6 A current dominates everywhere, since that wire is both closer and stronger. On the side of wire 2 (the weak one), however: $\frac{6}{x}=\frac{3}{x-0.20}$ gives $6(x-0.20)=3x$, so $x=0.40$ m from wire 1, that is 20 cm beyond wire 2. Check: $\frac{6}{0.40}=\frac{3}{0.20}=15$ A/m, so the two fields each equal $3.0\times 10^{-6}$ T there, in opposite directions.

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.

$n=\frac{1200}{0.40}=3000$ turns/m. $B=\mu_{0}nI=(4\pi\times 10^{-7})(3000)(2.5)\approx 9.4\times 10^{-3}$ T, hundreds of times more than a single loop carrying a comparable current: the solenoid is the economical way to build a strong, uniform field.

The density falls to $n=\frac{1200}{0.80}=1500$ turns/m and the field is halved: $B\approx 4.7\times 10^{-3}$ T. No: 1200 turns over 40 cm and 1200 turns over 80 cm are two different solenoids. Only the turn DENSITY $n$ counts: the field is built turn by turn, locally.

Take an infinite solenoid with interior field $B=\mu_{0}nI$ and cut it, in thought, into two semi-infinite halves. At the cut, each half creates the same field (they are mirror images of each other) and their sum rebuilds $B$: each therefore contributes exactly $\frac{B}{2}=\frac{\mu_{0}nI}{2}\approx 4.7\times 10^{-3}$ T. Now the end of a long solenoid is precisely a semi-infinite half seen from its tip: the field there is half the interior field. The field lines fan out at the ends, and on the axis the field drops to half its interior value.

Rotational symmetry requires that $\vec{B}$ be tangent to the circle of radius $r$ and of constant magnitude on it. The loop is therefore threaded by a field everywhere parallel to $d\vec{l}$: $\oint\vec{B}\cdot d\vec{l}=B(2\pi r)$. This circle encloses the $N$ turns, each once, so $I_{\text{enc}}=NI$ and $B=\frac{\mu_{0}NI}{2\pi r}$. In the central hole, a circle encloses NO turn: $I_{\text{enc}}=0$ and $B=0$. Outside the toroid, a circle encloses each turn twice, going and coming back, hence in opposite senses: $I_{\text{enc}}=NI-NI=0$ and $B=0$ again. The toroid confines its field entirely.

$B(0.080)=\frac{(2\times 10^{-7})(500)(3.0)}{0.080}=3.75\times 10^{-3}$ T; $B(0.10)=3.00\times 10^{-3}$ T; $B(0.12)=2.50\times 10^{-3}$ T. The field is therefore NOT uniform across the cross-section: it varies as $\frac{1}{r}$, from $+25$ % at the inner edge to $-17$ % at the outer edge relative to the mean radius, a total difference of 50 % between the two edges. The thinner the toroid is compared with its radius, the more negligible this variation becomes.

The number of turns per metre, measured along the mean circumference, is $n=\frac{N}{2\pi r_{mean}}=\frac{500}{2\pi(0.10)}\approx 796$ turns/m. Then $\mu_{0}nI=(4\pi\times 10^{-7})(796)(3.0)\approx 3.00\times 10^{-3}$ T: exactly $B(r_{mean})$. This is no coincidence, it is the same equation rewritten, since a toroid of large radius is locally indistinguishable from a straight solenoid. The advantage of the toroid lies elsewhere: because the field is strictly zero outside, all the flux stays in the core. A toroid does not radiate toward its neighbours and is not disturbed by them, whereas a straight solenoid lets its flux leak out of both ends. Hence its systematic use whenever inductance is wanted without crosstalk.

By symmetry, the field of an infinite current plane is parallel to the plane, perpendicular to the current, in opposite directions on either side, and of magnitude independent of the distance (nothing in the geometry provides a length scale). Rectangular loop of width $w$, straddling the plane, with two sides parallel to the field and two sides perpendicular to it: the perpendicular sides do not contribute, the other two each give $Bw$, that is $2Bw$ in total. Enclosed current: $J_{s}w$. Ampère: $2Bw=\mu_{0}J_{s}w$, so $B=\frac{\mu_{0}J_{s}}{2}=\frac{(4\pi\times 10^{-7})(500)}{2}\approx 3.14\times 10^{-4}$ T, the same value at 1 mm as at 1 km from the plane.

The two sheets each create $\frac{\mu_{0}J_{s}}{2}$. Between them, since the currents are opposite, the two fields point in the SAME direction and add: $B=\mu_{0}J_{s}\approx 6.28\times 10^{-4}$ T. Outside, they are opposite and cancel: $B=0$. This is exactly the structure of a solenoid, which is just a rolled-up current sheet: setting $J_{s}=nI$, we recover $B=\mu_{0}nI$ inside and zero outside. And it is also, point for point, the parallel-plate capacitor of electrostatics: one charged plane creates $\frac{\sigma}{2\varepsilon_{0}}$, two opposite planes give $\frac{\sigma}{\varepsilon_{0}}$ between them and zero outside. Two chapters, two constants, a single geometry: planar symmetry always produces a uniform field, confined between the two sources.