Worked examples — Magnetic field of straight wire, circular loop, solenoid, toroid

This page drills the parent topic until no case can surprise you. We first lay out every kind of question the four formulas can produce, then work an example for each cell — including zero cases, limiting cases, a real-world word problem, and an exam twist.

Everything below uses only four boxed results and two master laws (Biot-Savart Law, Ampere's Law). If a symbol appears, it was defined in the parent note; where a new one enters, I define it here first.

---

The scenario matrix

Read this as a checklist. Each row is a class of problem; the last column names the worked example that covers it. Note that Ex 4 covers two separate cells (D and E) — the same coil is used to illustrate the on-axis/far-field case and the degenerate centre case.

#	Geometry	Case class it tests	Covered by
A	Straight wire	Basic magnitude + direction (sign of field around wire)	Ex 1
B	Straight wire	Superposition of two wires — same vs opposite current sense	Ex 2
C	Straight wire	Finite wire, and the degenerate "point on the wire's line" ($\theta =0$)	Ex 3
D	Circular loop	Centre vs on-axis; limiting case $x\gg a$ (far field)	Ex 4 (part b, c)
E	Circular loop	Zero/degenerate point ($x=0$ recovered) + $N$-turn coil	Ex 4 (part a)
F	Solenoid	Interior field, end ($\frac{1}{2}$), radius-independence	Ex 5
G	Toroid	Inner edge vs outer edge (the $1/r$ variation), outside = 0	Ex 6
H	Real-world	Word problem: sizing a coil to hit a target field	Ex 7
I	Exam twist	Combining loop + straight wire; when fields can and cannot cancel	Ex 8

Throughout, ${\mu }_0=4\pi \times 10^{-7}\text{T⋅m/A}$, so the handy combo $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}\text{T⋅m/A}$.

---

Ex 1 — Straight wire, magnitude + direction (Cell A)

Forecast: Guess the order of magnitude — micro-tesla? milli-tesla? And point your right thumb up: which way do your fingers sweep on your east side — toward you or away?

1. Pick the formula. Infinite straight wire → $B=\frac{{\mu }_{0}I}{2\pi r}$. Why this step? One wire, full circular symmetry ⇒ Ampère's law already did the integral for us in the parent note.
2. Convert units. $r=5\text{cm}=0.05\text{m}$. Why this step? ${\mu }_0$ is in metres; mixing cm gives a $100\times$ error.
3. Plug in. $B=\frac{(4\pi \times 10^{-7})(10)}{2\pi (0.05)}=4\times 10^{-5}\text{T}=40\mu \text{T}.$ Why this step? The $4\pi /2\pi =2$ cancels cleanly.
4. Direction (right-hand grip). Thumb points up (current up), fingers curl counter-clockwise seen from above. On your east side the fingers point north. Why this step? Direction is not optional — half of a field answer is the direction.

Verify: Units: $\frac{(\text{T⋅m/A})(\text{A})}{\text{m}}=\text{T}$ ✓. Magnitude $40\mu \text{T}$ is comparable to Earth's field ($\sim 50\mu \text{T}$) — sensible for 10 A at a hand's width.

---

Ex 2 — Two parallel wires, superposition (Cell B)

Forecast: In which case does the midpoint field vanish? Trust your right hand before reading.

1. Field of each wire at the midpoint. Each is $r=d/2=0.10\text{m}$ away: $B_1=B_2=\frac{{\mu }_{0}I}{2\pi r}=\frac{(2\times 10^{-7})(6)}{0.10}=1.2\times 10^{-5}\text{T}.$ Why this step? Superposition: total field = vector sum of each wire's field. First get each piece.
2. Same direction (a) — track each field's direction explicitly. Let both currents point up, out of the page-plane. Apply the grip rule to each wire at the midpoint:
   • The left wire's field there circulates and points upward (call it $+y$): ${\vec{B}}_1=+B_1\hat{y}$.
   • The right wire's field there points downward ($-y$): ${\vec{B}}_2=-B_2\hat{y}$. Because the midpoint sits on opposite sides of the two wires, the same counter-clockwise circulation gives opposite local directions. Adding the vectors: ${\vec{B}}_{\text{net}}=B_1\hat{y}-B_2\hat{y}=0.$ Why this step? The subtraction is not arbitrary — it comes from the two field vectors literally pointing along $+\hat{y}$ and $-\hat{y}$. Same magnitudes, opposite directions ⇒ genuine zero (matrix cell "net zero").
3. Opposite direction (b). Flip the right wire's current to point down. That flips its field at the midpoint from $-\hat{y}$ to $+\hat{y}$, so now ${\vec{B}}_2=+B_2\hat{y}$ and the two align: ${\vec{B}}_{\text{net}}=B_1\hat{y}+B_2\hat{y}=2.4\times 10^{-5}\text{T}(\text{along }+\hat{y})=24\mu \text{T}.$ Why this step? Reversing one current reverses its field direction, turning the cancelling subtraction into a reinforcing addition.

Verify: Sanity: for the same-direction case the two wires' fields must cancel by symmetry (the midpoint sits equidistant with opposite local field directions) — and they do. See Magnetic force on a current-carrying conductor for why same-direction currents nonetheless attract each other even though the midpoint field is zero.

---

Ex 3 — Finite wire and the degenerate on-line point (Cell C)

Forecast: Should a finite wire give more or less field than an infinite one at the same $r$? Guess the direction of the inequality.

New tool — the finite-wire result from Biot-Savart Law: $B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2),$ where ${\theta }_1,{\theta }_2$ are the angles the two ends make, measured from the perpendicular dropped onto the point. Why these angles? They record how much of the wire "wraps around" the point; an infinite wire fills both to $90^{\circ }$. Note the constant here is $\frac{{\mu }_0}{4\pi }=10^{-7}$ (not $\frac{{\mu }_0}{2\pi }$) — we will use $\frac{{\mu }_0}{4\pi }$ consistently throughout this example.

1. Find the angles. Symmetric point ⇒ ${\theta }_1={\theta }_2=\theta$ with $\sin \theta =\frac{L}{\sqrt{L^2+r^2}}=\frac{4}{\sqrt{4^2+3^2}}=\frac{4}{5}=\mathrm{0.8.}$ Why this step? $\sin \theta$ = (half-length)/(distance to end); pure geometry of the right triangle end–foot–point.
2. Plug in — one consistent constant. Using $\frac{{\mu }_0}{4\pi }=10^{-7}$ throughout: $\frac{{\mu }_{0}I}{4\pi r}=\frac{(10^{-7})(8)}{0.03}=2.667\times 10^{-5}\text{T}.$ Then multiply by $(\sin {\theta }_1+\sin {\theta }_2)=0.8+0.8=1.6$: $B=(2.667\times 10^{-5})(1.6)=4.27\times 10^{-5}\text{T}=42.7\mu \text{T}.$ Why this step? Biot–Savart's finite-wire formula carries the $\frac{{\mu }_0}{4\pi }$ constant; we keep it fixed to avoid mixing factors of 2.
3. Compare to infinite wire. The infinite-wire formula uses the other combo $\frac{{\mu }_0}{2\pi }=2\times 10^{-7}$: $B_{\infty }=\frac{{\mu }_{0}I}{2\pi r}=\frac{(2\times 10^{-7})(8)}{0.03}=5.33\times 10^{-5}\text{T}.$ The finite wire gives less ($42.7<53.3\mu \text{T}$), because $\sin \theta <1$. Why this step? Confirms the forecast: a shorter wire has less "wrap," so weaker field. (The two formulas use different constants only because one has $4\pi$ and the other $2\pi$ baked in; they agree in the limit — see Verify.)
4. Degenerate point on the line. If the field point lies on the wire's extension, then $d\vec{l}$ and $\hat{r}$ are parallel, so $d\vec{l}\times \hat{r}=0$ ⇒ $B=0$. Why this step? The cross product kills contributions from a point staring straight down the wire — a case students forget.

Verify: Limit check: let $L\to \infty$ ⇒ $\sin \theta \to 1$ ⇒ $B\to \frac{{\mu }_{0}I}{4\pi r}(1+1)=\frac{{\mu }_{0}I}{2\pi r}$, recovering the infinite wire — the two constants reconcile exactly. ✓

---

Ex 4 — Loop: centre, on-axis, far-field limit, $N$ turns (Cells D & E)

Forecast: Will the centre or the $x=12\text{cm}$ point be stronger? By how rough a factor?

1. Centre ($x=0$), $N$ turns — this is Cell E. $B=\frac{{\mu }_{0}NI}{2a}=\frac{(4\pi \times 10^{-7})(100)(2)}{2(0.05)}=\frac{2.513\times 10^{-4}}{0.1}=2.51\times 10^{-3}\text{T}.$ Why this step? $x=0$ collapses the axis formula to ${\mu }_{0}NI/2a$ (the degenerate on-axis point). Multiply by $N$ for turns.
2. On-axis at $x=12\text{cm}$ — this is Cell D. $B=\frac{{\mu }_{0}NI{a}^2}{2(a^2+x^2{)}^{3/2}}.$ Here $a^2+x^2=0.05^2+0.12^2=0.0025+0.0144=0.0169$, so $(a^2+x^2{)}^{3/2}=0.0169^{1.5}=2.197\times 10^{-3}$. $B=\frac{(4\pi \times 10^{-7})(100)(2)(0.0025)}{2(2.197\times 10^{-3})}=\frac{6.283\times 10^{-7}}{4.394\times 10^{-3}}=1.43\times 10^{-4}\text{T}.$ Why this step? Off the centre we must use Biot–Savart's axial result — no Amperian shortcut exists here.
3. Far-field limit — also Cell D. When $x\gg a$, drop $a^2$ under the root: $(a^2+x^2{)}^{3/2}\to x^3$, giving $B\approx \frac{{\mu }_{0}NI{a}^2}{2x^3}\propto \frac{1}{x^3}.$ Why this step? This $1/x^3$ is the fingerprint of a magnetic dipole — the loop is one from far away.

Verify: Ratio check: $\frac{B_{\text{centre}}}{B_{x=12}}=\frac{(a^2+x^2{)}^{3/2}}{a^3}=\frac{2.197\times 10^{-3}}{1.25\times 10^{-4}}=17.6$, and directly $2.51\times 10^{-3}/1.43\times 10^{-4}\approx 17.6$ ✓.

---

Ex 5 — Solenoid: interior, end, radius-independence (Cell F)

Forecast: Which is bigger — the deep-inside field or the end field? By what factor exactly?

1. Turns per metre. $n=N/L=1000/0.5=2000{\text{m}}^{-1}$. Why this step? The solenoid formula $B={\mu }_{0}nI$ needs $n$, not $N$.
2. Interior. $B={\mu }_{0}nI=(4\pi \times 10^{-7})(2000)(3)=7.54\times 10^{-3}\text{T}.$ Why this step? Ampère's rectangular loop gave a field independent of position inside — this single number holds everywhere in the bulk.
3. End of a (semi-infinite) solenoid. $B_{\text{end}}=\frac{1}{2}{\mu }_{0}nI=3.77\times 10^{-3}\text{T}.$ Why this step? At the mouth, only "half" the solenoid lies on each side, so exactly half the field — a limiting/degenerate boundary case.
4. Double the radius. $B$ unchanged: ${\mu }_{0}nI$ contains no $r$. Why this step? The formula has no radius in it, so widening the coil (at fixed $n$ and $I$) does not change the bulk field — the field depends only on how tightly the turns are packed, not on the tube's width.

Verify: Ratio $B_{\text{interior}}/B_{\text{end}}=2$ exactly ✓. Units: $(\text{T⋅m/A})({\text{m}}^{-1})(\text{A})=\text{T}$ ✓.

---

Ex 6 — Toroid: inner vs outer edge, outside zero (Cell G)

Forecast: Where is the field strongest — inner or outer edge? Recall $B\propto 1/r$.

1. Inner edge. $B_{\text{in}}=\frac{{\mu }_{0}NI}{2\pi r_{\text{in}}}=\frac{(2\times 10^{-7})(500)(4)}{0.08}=5.0\times 10^{-3}\text{T}.$ Why this step? Ampère's circular loop inside the core encloses all $N$ turns; smaller $r$ ⇒ larger $B$.
2. Outer edge. $B_{\text{out}}=\frac{{\mu }_{0}NI}{2\pi r_{\text{out}}}=\frac{(2\times 10^{-7})(500)(4)}{0.10}=4.0\times 10^{-3}\text{T}.$ Why this step? Same enclosed current, larger $r$ ⇒ weaker field — the $1/r$ non-uniformity in action.
3. Just outside the toroid. An Amperian circle outside the core encloses zero net current (each turn crosses it twice, in and out) ⇒ $B=0$. Why this step? This zero is the toroid's signature — the field is trapped inside.

Verify: Ratio $B_{\text{in}}/B_{\text{out}}=r_{\text{out}}/r_{\text{in}}=0.10/0.08=1.25$, and $5.0/4.0=1.25$ ✓. Inner is indeed stronger, confirming the forecast.

---

Ex 7 — Real-world word problem (Cell H)

Forecast: Will the required current be "a few amps" or "hundreds of amps"? Guess before computing.

1. Solve $B={\mu }_{0}nI$ for $I$. $I=\frac{B}{{\mu }_{0}n}=\frac{0.01}{(4\pi \times 10^{-7})(4000)}=\frac{0.01}{5.027\times 10^{-3}}=1.99\text{A}.$ Why this step? The demo fixes $B$; we invert the solenoid formula to get the unknown current.
2. Supply-limited version. With $I=2\text{A}$ capped, solve for $n$: $n=\frac{B}{{\mu }_{0}I}=\frac{0.01}{(4\pi \times 10^{-7})(2)}=3.98\times 10^3{\text{m}}^{-1}\approx 4000\text{turns/m}.$ Why this step? Now $I$ is fixed and $n$ is the design knob — same equation, different unknown.

Verify: Both answers land at essentially the same operating point ($\approx 2\text{A}$, $\approx 4000\text{turns/m}$), a self-consistency check. A ~2 A demo current is very reasonable — matches the forecast "a few amps." The uniform field produced could then drive a demo via flux through a pickup coil (Faraday's law of induction).

---

Ex 8 — Exam twist: loop + wire, when fields cancel (Cell I)

Forecast: The wire's field circles around the axis; the loop's centre field points along it. Are these two fields even parallel? That single fact decides whether cancellation is possible — guess now.

1. Loop centre field (its magnitude and direction). $B_L=\frac{{\mu }_0{I}_L}{2a}=\frac{(4\pi \times 10^{-7})(3)}{2(0.04)}=4.71\times 10^{-5}\text{T},\text{directed along the axis (out of page).}$ Why this step? Establish the loop's contribution and, crucially, its direction — it is axial.
2. Wire current that matches that magnitude. Set the straight-wire field equal to $B_L$: $\frac{{\mu }_0{I}_W}{2\pi r}=B_L\Rightarrow I_W=\frac{B_{L}2\pi r}{{\mu }_0}=\frac{B_{L}r}{2\times 10^{-7}}=\frac{(4.71\times 10^{-5})(0.02)}{2\times 10^{-7}}=4.71\text{A}.$ Why this step? Invert the straight-wire law for the current that reproduces the target magnitude. (Numerically the same digits as $I_L$ appear here only by coincidence of the chosen numbers.)
3. Can they cancel? — the trap, resolved by geometry. The wire's field is perpendicular to the wire: it circles around the axis, so it lies in the loop's plane. The loop's centre field points along the axis. Two vectors at $90^{\circ }$ can never be anti-parallel, so they cannot cancel. The net magnitude is $B_{\text{net}}=\sqrt{B_L^2+B_W^2}=\sqrt{2}{B}_L=6.66\times 10^{-5}\text{T},$ directed at $45^{\circ }$ to the axis — never zero. Why this step? The exam bait is "make them cancel." Cancellation requires anti-parallel vectors; here they are orthogonal, so the answer is a firm "no matter what $I_W$ is, the net is non-zero."

Verify: With $I_W=4.71\text{A}$ the two magnitudes are equal, so the net is $\sqrt{2}{B}_L=6.66\times 10^{-5}\text{T}$ at $45^{\circ }$ to the axis — decidedly non-zero, confirming step 3.

---

Recall Rapid self-test (cover the right side)

Field of infinite wire, 10 A at 5 cm ::: $40\mu \text{T}$ Midpoint of two 6 A wires, same direction, 20 cm apart ::: $0$ (fields point opposite, cancel) Same, opposite direction ::: $24\mu \text{T}$ Point lying on a wire's own line ::: $B=0$ (since $d\vec{l}\parallel \hat{r}$) Loop $1/x^3$ far-field means the loop behaves like a ::: magnetic dipole Field just outside a toroid ::: $0$ (zero enclosed current) Can a wire-on-axis field cancel a loop's centre field? ::: No — they are perpendicular