Exercises — Magnetic field of straight wire, circular loop, solenoid, toroid

WHAT tool: infinite straight wire → $B=\frac{{\mu }_{0}I}{2\pi r}$ (Ampère's law result, chosen because the wire has circular symmetry). WHY: we are given current and distance only — this is the textbook straight-wire case. Convert $r=5\text{cm}=0.05\text{m}$. $B=\frac{(4\pi \times 10^{-7})(10)}{2\pi (0.05)}=\frac{4\pi \times 10^{-6}}{0.1\pi }=4\times 10^{-5}\text{T}$ Direction: grip rule — field circles the wire. Answer: $B=4\times 10^{-5}\text{T}=40\mu \text{T}$.

WHAT tool: $B=\frac{{\mu }_{0}I}{2a}$ — the centre-of-loop special case ($x=0$). WHY: field is asked at the exact centre, so we do not need the full axial formula. $a=0.04\text{m}$. $B=\frac{(4\pi \times 10^{-7})(3)}{2(0.04)}=\frac{12\pi \times 10^{-7}}{0.08}=1.5\pi \times 10^{-5}\approx 4.71\times 10^{-5}\text{T}$ Direction: right-hand curl rule — thumb along the axis gives $\vec{B}$ perpendicular to the loop plane. Answer: $B\approx 4.71\times 10^{-5}\text{T}$.

WHAT tool: $B={\mu }_{0}nI$ — "naked" formula, no radius needed. WHY: "deep inside a long solenoid" is exactly where the ideal formula applies. $B=(4\pi \times 10^{-7})(500)(2)=4\pi \times 10^{-4}\approx 1.257\times 10^{-3}\text{T}$ Direction: along the axis, sense given by curling fingers with the current. Answer: $B\approx 1.26\times 10^{-3}\text{T}=1.26\text{mT}$.

WHAT tool: $B=\frac{{\mu }_{0}I{a}^2}{2(a^2+x^2{)}^{3/2}}$ — the full axial formula (Biot–Savart). WHY this tool: the field point sits off the centre ($x=0.10\text{m}\ne 0$), so the simple ${\mu }_{0}I/2a$ centre formula no longer applies. And why can't Ampère's law shortcut it? Ampère's law only hands you $B$ when you can find a closed path along which $B$ has the same magnitude everywhere and points along the path — then $\oint \vec{B}\cdot d\vec{l}=B\times (\text{length})$. On the axis of a single loop, the field magnitude changes from point to point and the field lines curve away from any circle or line you might draw, so no such "nice" path exists. The equation $\oint \vec{B}\cdot d\vec{l}={\mu }_0{I}_{\text{enc}}$ is still true, it just can't be solved for $B$. So we fall back to integrating Biot–Savart element by element. Look at the figure below: only the axial (green) components of each element's field survive; the sideways parts cancel in pairs.

Here $a=x=0.10$, so $a^2+x^2=0.02{\text{m}}^2$ and $(a^2+x^2{)}^{3/2}=(0.02{)}^{1.5}$. $B=\frac{(4\pi \times 10^{-7})(5)(0.01)}{2(0.02{)}^{3/2}}$ Now $(0.02{)}^{3/2}=0.02\sqrt{0.02}=0.02\times 0.14142=2.828\times 10^{-3}$. $B=\frac{(4\pi \times 10^{-7})(5)(0.01)}{2(2.828\times 10^{-3})}=\frac{2\pi \times 10^{-8}}{5.657\times 10^{-3}}\approx 1.111\times 10^{-5}\text{T}$ Centre value: $B_0=\frac{{\mu }_{0}I}{2a}=\frac{(4\pi \times 10^{-7})(5)}{0.2}\approx 3.14\times 10^{-5}\text{T}$. Ratio: $B/B_0=a^3/(a^2+x^2{)}^{3/2}=(0.10{)}^3/(0.02{)}^{1.5}=10^{-3}/2.828\times 10^{-3}\approx 0.354$. Direction: the surviving field (the green axial arrow in Figure s01) points along the axis, its sense set by the right-hand curl rule — curl your fingers with the loop's current, and your thumb gives the direction of $\vec{B}$ along the axis (the same direction as at the centre, just weaker). Answer: $B\approx 1.11\times 10^{-5}\text{T}$ directed along the axis, about 35% of the centre value.

WHAT tool: $B=\frac{{\mu }_{0}NI}{2\pi r}$ (Ampère's law, circular symmetry inside the doughnut). WHY: the field inside a toroid depends on $r$; along the mean circle we use the mean radius. $B=\frac{(4\pi \times 10^{-7})(1200)(4)}{2\pi (0.15)}=\frac{4\pi \times 10^{-7}\times 4800}{0.3\pi }=\frac{1.92\times 10^{-3}\pi }{0.3\pi }$ WHY cancel the $\pi$: the same factor $\pi$ appears in both the top and the bottom of the fraction, so it divides out to $1$ — a shared factor top-and-bottom never changes the value, it just makes the arithmetic cleaner. What remains is a plain division: $B=\frac{1.92\times 10^{-3}}{0.3}=6.4\times 10^{-3}\text{T}$ Direction: the field runs circularly inside the core, sense set by curling fingers with the winding current. Answer: $B=6.4\times 10^{-3}\text{T}=6.4\text{mT}$.

First, the idealization warning. The clean result $B=\frac{1}{2}{\mu }_{0}nI$ is exact only for an ideal semi-infinite solenoid — infinitely long on one side, thin walls, tightly wound. A real finite solenoid's end field lies close to but not exactly half of ${\mu }_{0}nI$: near the middle of a long-but-finite coil the field is $\approx {\mu }_{0}nI$, and it tapers down to roughly one-half at each mouth, then keeps decreasing outside. So use "half at the end" as an excellent estimate for a coil that is much longer than it is wide, and expect small corrections for stubby coils. WHAT tool: at the end of a semi-infinite solenoid, $B=\frac{1}{2}{\mu }_{0}nI$ — exactly half the deep-inside value. WHY exactly half (short derivation): think of a solenoid as a stack of current loops. Deep inside, loops extend to both sides of the field point, and their axial contributions add up to ${\mu }_{0}nI$. Model this as an integral of loop-fields over all angles: an infinite solenoid collects contributions from $\theta =0$ to $\theta =180°$ (loops on both sides), giving the full ${\mu }_{0}nI$. At the open mouth of a semi-infinite solenoid, the winding only exists on one side of the field point, so the integral runs over just half the angular range ($\theta =90°$ to $180°$). By the symmetry of that integral, each half contributes equally, so the end value is precisely $\frac{1}{2}{\mu }_{0}nI$. $B=\frac{1}{2}(4\pi \times 10^{-7})(500)(2)=\frac{1}{2}(1.257\times 10^{-3})\approx 6.28\times 10^{-4}\text{T}$ Answer: $B\approx 6.28\times 10^{-4}\text{T}=0.628\text{mT}$.

WHAT tool: finite-wire Biot–Savart result $B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)$, where ${\theta }_1,{\theta }_2$ are the angles each wire end subtends measured from the perpendicular foot. WHY this tool (not the ${\mu }_{0}I/2\pi r$ shortcut): the ${\mu }_{0}I/2\pi r$ formula assumes an infinite wire so that Ampère's law can pull $B$ out of a circular loop. A finite wire breaks that symmetry (the two ends are at finite angles), so we must go back to Biot–Savart and integrate each element's contribution — the integral yields the $(\sin {\theta }_1+\sin {\theta }_2)$ factor, which is derived in full in the parent note. Setting both angles to $90°$ recovers the infinite-wire result, confirming the two agree in the limit. Look at the geometry figure:

Tying the algebra to the figure: in Figure s02, the dashed line of length $r$ is the perpendicular from P to the wire; the far end sits a distance $\ell$ along the wire from that foot. Those two legs ($r$ across, $\ell$ along) form a right triangle whose hypotenuse is $\sqrt{{\ell }^2+r^2}$, the straight-line distance from P to the wire's end. The angle $\theta$ in the formula is the angle at P between the perpendicular and that hypotenuse, so its sine = opposite / hypotenuse = $\ell /\sqrt{{\ell }^2+r^2}$ — this is exactly the $\sin \theta$ the Biot–Savart integral asks for. Because P sits opposite the midpoint, the triangle on the left mirrors the one on the right, so ${\theta }_1={\theta }_2=\theta$. $\sin \theta =\frac{\ell }{\sqrt{{\ell }^2+r^2}}=\frac{0.12}{\sqrt{0.12^2+0.05^2}}=\frac{0.12}{\sqrt{0.0144+0.0025}}=\frac{0.12}{0.13}\approx 0.9231$ $B=\frac{{\mu }_{0}I}{4\pi r}(2\sin \theta )=\frac{(4\pi \times 10^{-7})(6)}{4\pi (0.05)}(2\times 0.9231)$ $=\frac{6\times 10^{-7}}{0.05}(1.8462)=(1.2\times 10^{-5})(1.8462)\approx 2.215\times 10^{-5}\text{T}$ Answer: $B\approx 2.22\times 10^{-5}\text{T}$. Notice it is less than the infinite-wire value $\frac{{\mu }_{0}I}{2\pi r}=2.4\times 10^{-5}\text{T}$, because a finite wire has fewer current elements contributing. ✓

WHAT tool: superposition — add each wire's field vectorially. At the midpoint, each wire is $r=0.10\text{m}$ away. WHY vectors matter: for currents in the same direction, at the midpoint the two fields point in opposite directions (grip rule), so they subtract. $B_A=\frac{{\mu }_0{I}_A}{2\pi r}=\frac{(4\pi \times 10^{-7})(8)}{2\pi (0.10)}=1.6\times 10^{-5}\text{T}$ $B_B=\frac{{\mu }_0{I}_B}{2\pi r}=\frac{(4\pi \times 10^{-7})(12)}{2\pi (0.10)}=2.4\times 10^{-5}\text{T}$ Opposite directions → net magnitude: $B_{\text{net}}=|B_B-B_A|=(2.4-1.6)\times 10^{-5}=0.8\times 10^{-5}\text{T}=8\times 10^{-6}\text{T}$ Answer: $B_{\text{net}}=8\times 10^{-6}\text{T}$, directed the way the stronger wire (B) dictates. (See Magnetic force on a current-carrying conductor for what this net field then does to a third wire.)

Step 1 — WHAT/WHY find $N$: each turn is one circumference of the rod, length $2\pi \rho$. So the number of turns is total wire length ÷ circumference: $N=\frac{L_{\text{wire}}}{2\pi \rho }=\frac{20}{2\pi (0.01)}=\frac{20}{0.0628}\approx 318.3\text{turns}$ Step 2 — WHAT/WHY turns per metre: the solenoid formula needs $n$ = turns packed per metre of length, so divide the turn count by the coil length ${\ell }_s$: $n=\frac{N}{{\ell }_s}=\frac{318.3}{0.25}\approx 1273\text{turns/m}$ Step 3 — field: now apply $B={\mu }_{0}nI$: $B={\mu }_{0}nI=(4\pi \times 10^{-7})(1273)(3)\approx 4.80\times 10^{-3}\text{T}$ Answer: $B\approx 4.80\times 10^{-3}\text{T}=4.80\text{mT}$. Insight: the rod radius $\rho$ entered only through how many turns fit — it does not appear in the final ${\mu }_{0}nI$. Radius affects $N$, not the field-per-turn.

Set up the equation (this is the synthesis: two different formulas set equal): $\frac{{\mu }_{0}I{a}^2}{2(a^2+x^2{)}^{3/2}}=\frac{{\mu }_{0}I}{2\pi a}$ Cancel ${\mu }_{0}I/2$ from both sides: $\frac{a^2}{(a^2+x^2{)}^{3/2}}=\frac{1}{\pi a}$ Cross-multiply: $\pi a\cdot a^2=(a^2+x^2{)}^{3/2}$, i.e. $\pi a^3=(a^2+x^2{)}^{3/2}$. Raise both sides to the $2/3$ power: $(\pi a^3{)}^{2/3}=a^2+x^2\Rightarrow {\pi }^{2/3}{a}^2=a^2+x^2$ $x^2=a^2({\pi }^{2/3}-1)\Rightarrow x=a\sqrt{{\pi }^{2/3}-1}$ Numerically ${\pi }^{2/3}=2.145$, so $x=a\sqrt{1.145}\approx 1.070a$. Answer: $x\approx 1.07a$ — just past one radius out along the axis.

(a) WHAT tool: $B=\frac{{\mu }_{0}NI}{2\pi r}$, evaluated at each radius (field is not uniform — it falls as $1/r$). First simplify the common numerator. ${\mu }_{0}NI=(4\pi \times 10^{-7})(2000)(5)=4\pi \times 10^{-3}$. Dividing by $2\pi$ gives $\frac{4\pi \times 10^{-3}}{2\pi }=2\times 10^{-3}$. So $B(r)=\frac{2\times 10^{-3}}{r}$. $B_{\text{inner}}=\frac{2\times 10^{-3}}{0.10}=2.0\times 10^{-2}\text{T}=20\text{mT}$ $B_{\text{outer}}=\frac{2\times 10^{-3}}{0.14}\approx 1.43\times 10^{-2}\text{T}=14.3\text{mT}$ (b) Percentage variation (relative to the inner, larger value): $\frac{B_{\text{inner}}-B_{\text{outer}}}{B_{\text{inner}}}=\frac{20-14.29}{20}\approx 0.2857=28.6\%$ Answer: inner $=20\text{mT}$, outer $\approx 14.3\text{mT}$; field drops about 28.6% across the core.

WHY Ampère's law: cylindrical symmetry lets $B$ leave the integral, so $\oint \vec{B}\cdot d\vec{l}=B(2\pi r)={\mu }_0{I}_{\text{enc}}$. The twist is that inside the wire only part of the current is enclosed.

Look at the figure — the enclosed current grows with the area of the Amperian circle:

(a) Inside ($r<R$): current density is uniform, so enclosed current scales as the area ratio: $I_{\text{enc}}=I\frac{\pi r^2}{\pi R^2}=I\frac{r^2}{R^2}$ $B(2\pi r)={\mu }_{0}I\frac{r^2}{R^2}\Rightarrow B=\frac{{\mu }_{0}Ir}{2\pi R^2}$ At $r=0.01$, $R=0.02$: $B=\frac{(4\pi \times 10^{-7})(30)(0.01)}{2\pi (0.02{)}^2}=\frac{(4\pi \times 10^{-7})(0.3)}{2\pi (4\times 10^{-4})}=\frac{1.2\pi \times 10^{-7}}{8\pi \times 10^{-4}}=1.5\times 10^{-4}\text{T}$

(b) Outside ($r>R$): now the whole current threads the loop, $I_{\text{enc}}=I$, so we recover the ordinary straight-wire formula: $B(2\pi r)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi r}$ At $r=0.04$: $B=\frac{(4\pi \times 10^{-7})(30)}{2\pi (0.04)}=\frac{1.2\pi \times 10^{-5}}{0.08\pi }=1.5\times 10^{-4}\text{T}$

Answer: both give $1.5\times 10^{-4}\text{T}$. This numeric coincidence happens because $r_{\text{inside}}=0.01$ and $r_{\text{outside}}=0.04$ were chosen to land on the same value. In general, inside $B\propto r$ (rising linearly from zero at the centre), and outside $B\propto 1/r$ (falling), with the maximum at the surface $r=R$: there $B_{\max }=\frac{{\mu }_{0}I}{2\pi R}=\frac{(4\pi \times 10^{-7})(30)}{2\pi (0.02)}=3.0\times 10^{-4}\text{T}$.