Biot-Savart law — magnetic field from current element: Worked examples

Everything here uses only three physical objects, so let us name them once in plain words before any symbol appears:

The two master formulas we will lean on (both derived in the parent note):

We will justify the arc formula the first time we use it (Example E). And $μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A$ throughout.

The scenario matrix

Every straight-or-circular Biot–Savart problem falls into one of these cells. The last column names the worked example that covers it.

#	Case class	The tricky bit	Worked in
C1	Infinite wire, both ends far away	$θ_{1} = θ_{2} = 9 0^{\circ}$	A
C2	Semi-infinite wire, one end AT the foot	one angle $= 0^{\circ}$	B
C3	Finite wire, $P$ off to the side (asymmetric)	two different non- $9 0^{\circ}$ angles, sign of angle	C
C4	Point ON the line of the wire (degenerate)	$sin θ = 0$ everywhere ⇒ $B = 0$	D
C5	Circular arc (partial loop)	fraction $ϕ /2 π$ of a full loop	E
C6	Superposition — two straight pieces + an arc (a bent conductor)	add vectors, watch directions	F
C7	On-axis point of a loop, limiting behaviour	far-field $\to 1/ x^{3}$ dipole	G
C8	Real-world word problem with numbers & units	unit bookkeeping, order of magnitude	H

We now clear every cell.

Worked example Example A — Cell C1: Infinite straight wire

A very long straight wire carries $I = 10 A$ . Find $B$ at $a = 5 cm = 0.05 m$ .

Forecast: Before computing — will $B$ be closer to $1 0^{- 5} T$ or $1 0^{- 2} T$ ? (Earth's field is $\sim 5 \times 1 0^{- 5} T$ ; guess now.)

Identify the angles. Both ends run off to infinity, so from $P$ each end is seen at $9 0^{\circ}$ from the perpendicular. Why this step? The workhorse formula needs $θ_{1}, θ_{2}$ ; infinite ends both give the extreme angle $9 0^{\circ}$ .
Plug in $sin 9 0^{\circ} = 1$ : $B = \frac{μ _{0} I}{4 π a} (1 + 1) = \frac{μ _{0} I}{2 π a}$ . Why? The two $1$ s add to $2$ , collapsing the finite formula into the famous infinite one.
Numbers: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 10 )}{2 π ( 0.05 )} = 4 \times 1 0^{- 5} T$ .

Verify: Units: $\frac{( T\cdotm/A ) ( A )}{m} = T$ . ✔ Magnitude $4 \times 1 0^{- 5} T$ — comparable to Earth's field, so your compass would notice. The $μ_{0} = 4 π \times 1 0^{- 7}$ cancelled the $2 π$ neatly, a good sign.

Worked example Example B — Cell C2: Semi-infinite wire

A straight wire starts at the perpendicular foot $O$ and runs to $+ \infty$ in one direction only (imagine a wire coming up out of a wall). Field point $P$ is at distance $a$ from $O$ . Show $B$ is exactly half the infinite-wire value.

Forecast: Half a wire ⇒ half the field? Guess yes/no before the algebra.

Angle at the near end. The wire's near end sits at the foot $O$ , straight down from $P$ . Looking from $P$ toward $O$ , the line lies along the perpendicular itself, so $θ_{1} = 0^{\circ}$ . Why this step? Look at the figure: the angle from the perpendicular foot to the near end is zero because the end is the foot.
Angle at the far end. The far end runs to infinity, so $θ_{2} = 9 0^{\circ}$ . Why? Same reasoning as the infinite case for that side.
Plug in: $B = \frac{μ _{0} I}{4 π a} (sin 0^{\circ} + sin 9 0^{\circ}) = \frac{μ _{0} I}{4 π a} (0 + 1) = \frac{μ _{0} I}{4 π a}$ .

Verify: Compare to infinite value $\frac{μ _{0} I}{2 π a}$ : the ratio is exactly $\frac{1}{2}$ . ✔ So a semi-infinite wire gives half — the forecast "half" was right, but note why: the far half contributes the full $9 0^{\circ}$ side and the near end contributes $0$ , not because you literally sliced the wire in two equal field-halves.

Worked example Example C — Cell C3: Finite wire,

$P$ off to one side (asymmetric) A straight wire runs from $y = 0$ up to $y = L$ along the $y$ -axis, $I = 8 A$ . The field point $P$ is at perpendicular distance $a = 3 m$ , level with the bottom end ( $y = 0$ ), $L = 4 m$ . Find $B$ .

Forecast: Will the two end-angles be equal or different here? Which sign issue could bite you?

Where is the perpendicular foot? $P$ is level with the bottom end, so the foot $O$ is at the bottom end $(y = 0)$ . Why this step? The formula's angles are measured from the foot; we must locate it first. Here it coincides with an end.
Angle to the bottom end: it is at the foot, so $θ_{1} = 0^{\circ}$ , $sin θ_{1} = 0$ . Why? Same degeneracy as Example B's near end.
Angle to the top end: the top end is at height $L = 4$ , horizontal distance $a = 3$ . The angle from the perpendicular to the line $P \to top$ satisfies $sin θ_{2} = \frac{L}{a ^{2} + L ^{2}} = \frac{4}{9 + 16} = \frac{4}{5} = 0.8$ . Why this step? On the right triangle (legs $a$ and $L$ ), the side opposite the angle-at- $P$ is the vertical leg $L$ ; hence opposite/hypotenuse $= L / a^{2} + L^{2}$ . This is why we use $sin$ : it is opposite-over-hypotenuse on exactly this triangle.
Combine: $B = \frac{μ _{0} I}{4 π a} (0 + 0.8) = \frac{( 1 0 ^{- 7} ) ( 8 ) ( 0.8 )}{3}$ (using $μ_{0} /4 π = 1 0^{- 7}$ ).
Number: $B = \frac{8 \times 0.8 \times 1 0 ^{- 7}}{3} = 2.13 \times 1 0^{- 7} T$ .

Verify: Sanity — this must be less than the infinite-wire value at $a = 3$ , which is $\frac{μ _{0} I}{2 π a} = \frac{( 2 \times 1 0 ^{- 7} ) ( 8 )}{3} = 5.33 \times 1 0^{- 7} T$ . Our $2.13 \times 1 0^{- 7}$ is smaller ✔ (a stub of wire gives less than an infinite one). The sign trap: if $P$ were level between the two ends, one angle would be measured on the other side of the foot and both $sin$ values add as positives — always take each angle as a positive amount from the perpendicular and add.

Worked example Example D — Cell C4: Point ON the line of the wire (degenerate)

A straight wire lies along the $x$ -axis. The field point $P$ sits on the $x$ -axis too, 2 m beyond the wire's end. What is $B$ at $P$ ?

Forecast: Non-zero or exactly zero? Commit before reading.

Look at $θ$ for every element. For each element $d l$ (pointing along $x$ ) the vector $\overset{r}{^}$ from that element to $P$ also points along $x$ . So the angle between them is $θ = 0^{\circ}$ . Why this step? Biot–Savart's magnitude carries $sin θ$ ; we must inspect $θ$ before integrating.
Apply $sin θ$ : $sin 0^{\circ} = 0$ , so $d B = 0$ for every element. Why? A current element sprays field sideways only — never straight ahead or straight behind. $P$ is straight ahead of the whole wire.
Integrate: $B = \int 0 = 0$ .

Verify: $B = 0$ exactly. ✔ This is the "straight-ahead zero" fact from the parent note made literal: a point on the wire's own axis feels no field from that wire. Any tiny sideways offset breaks the tie and gives the finite-wire formula.

Worked example Example E — Cell C5: Circular arc (partial loop)

A wire is bent into a quarter circle (arc angle $9 0^{\circ}$ ) of radius $R = 0.2 m$ , current $I = 6 A$ . Find $B$ at the centre of the circle.

Forecast: A full loop gives $μ_{0} I /2 R$ . A quarter should give what fraction of that?

Every element is perpendicular to its $\overset{r}{^}$ . On a circle the tangent (direction of $d l$ ) is always at $9 0^{\circ}$ to the radius (direction of $\overset{r}{^}$ toward the centre). So $sin θ = 1$ and $r = R$ for all elements. Why this step? This is what makes the arc integral trivial — no varying $sin θ$ , no varying $r$ .
Write $d B$ and integrate arc length. $d B = \frac{μ _{0} I}{4 π R ^{2}} d l$ , and all point the same way (out of the plane, right-hand rule), so add magnitudes: $B = \frac{μ _{0} I}{4 π R ^{2}} \int d l$ . Why? Same direction ⇒ scalar sum, no cancellation.
Arc length of a $ϕ$ -radian arc is $R ϕ$ . For a quarter, $ϕ = π /2$ , so $\int d l = R \cdot \frac{π}{2}$ . Then $B = \frac{μ _{0} I}{4 π R ^{2}} \cdot R \frac{π}{2} = \frac{μ _{0} I}{8 R}$ . Why this step? This is exactly the general arc formula $B = \frac{μ _{0} I}{4 π R} ϕ$ evaluated at $ϕ = π /2$ .
Number: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 6 )}{8 ( 0.2 )} = 4.71 \times 1 0^{- 6} T$ .

Verify: Fraction check: quarter arc $= \frac{1}{4}$ of a full loop, and $\frac{μ _{0} I}{8 R} = \frac{1}{4} \cdot \frac{μ _{0} I}{2 R}$ . ✔ Forecast "one quarter" confirmed. Full-loop value would be $\frac{μ _{0} I}{2 R} = 1.885 \times 1 0^{- 5} T$ ; ours is a quarter of that.

Worked example Example F — Cell C6: Superposition (bent conductor)

A wire comes in from far away along a radius, bends into a half circle (radius $R = 0.1 m$ ) around a point $O$ , then leaves along another radius to infinity. Current $I = 5 A$ . Find $B$ at $O$ .

Forecast: The straight pieces — do they contribute at $O$ , or zero?

The two straight radial pieces. Each straight piece points directly toward or away from $O$ , so for every element on them $\overset{r}{^}$ (toward $O$ ) is parallel/antiparallel to $d l$ : $θ = 0^{\circ}$ or $18 0^{\circ}$ , $sin θ = 0$ . Contribution $= 0$ . Why this step? Same degeneracy as Example D — a wire aimed at the field point gives nothing there.
The half-circle arc. Use the arc formula with $ϕ = π$ : $B_{arc} = \frac{μ _{0} I}{4 π R} π = \frac{μ _{0} I}{4 R}$ . Why? Straights gave zero, so the total is just the semicircle.
Number: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 5 )}{4 ( 0.1 )} = 1.571 \times 1 0^{- 5} T$ .

Verify: A half loop should be half of $\frac{μ _{0} I}{2 R}$ : $\frac{1}{2} \cdot \frac{( 4 π \times 1 0 ^{- 7} ) ( 5 )}{2 ( 0.1 )} = \frac{1}{2} (3.14 \times 1 0^{- 5}) = 1.571 \times 1 0^{- 5}$ . ✔ Matches. The lesson: in a bent conductor, radial-in / radial-out legs are free — they never contribute at the arc's centre.

Worked example Example G — Cell C7: On-axis loop, limiting behaviour

A loop radius $R = 0.05 m$ , $I = 3 A$ . Compute $B$ (a) at the centre and (b) on the axis at $x = 0.20 m$ , and confirm the far-field falls like $1/ x^{3}$ .

Forecast: Which of the two points has the weaker field, and by roughly what factor?

The on-axis formula (from the parent note) is $B (x) = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$ .

Centre ( $x = 0$ ): $B (0) = \frac{μ _{0} I R ^{2}}{2 R ^{3}} = \frac{μ _{0} I}{2 R} = \frac{( 4 π \times 1 0 ^{- 7} ) ( 3 )}{2 ( 0.05 )} = 3.77 \times 1 0^{- 5} T$ . Why this step? Setting $x = 0$ must recover the loop-centre result — a built-in consistency check.
At $x = 0.20$ : $R^{2} + x^{2} = 0.0025 + 0.04 = 0.0425$ , $(0.0425)^{3/2} = 8.76 \times 1 0^{- 3}$ . Then $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 3 ) ( 0.0025 )}{2 ( 8.76 \times 1 0 ^{- 3} )} = 5.38 \times 1 0^{- 7} T$ . Why? Direct substitution; keep the $3/2$ power exact.
Far-field check. When $x ≫ R$ , $(R^{2} + x^{2})^{3/2} \approx x^{3}$ , so $B \approx \frac{μ _{0} I R ^{2}}{2 x ^{3}} \propto \frac{1}{x ^{3}}$ . Why this step? Dropping $R^{2}$ beside $x^{2}$ exposes the dipole $1/ x^{3}$ tail — see Magnetic Dipole Moment.

Verify: Ratio $B (0) / B (0.20) = 3.77 \times 1 0^{- 5} /5.38 \times 1 0^{- 7} \approx 70$ . Rough $1/ x^{3}$ scaling would predict a big drop for $x = 4 R$ , and indeed the exact factor is $\approx 70$ . ✔ Units of $B (0.20)$ : T (all SI in, T out).

Worked example Example H — Cell C8: Real-world word problem

A household appliance cable carries $I = 15 A$ . A magnetic compass sits $d = 2 cm$ from the (long, straight) cable. Is the field strong enough to visibly deflect the compass, given Earth's horizontal field is about $B_{E} = 2 \times 1 0^{- 5} T$ ?

Forecast: Guess: comparable to Earth's field, ten times bigger, or ten times smaller?

Model the cable as an infinite straight wire. $2 cm$ is tiny next to a metre-long cable, so both ends look "infinite." Why this step? Justifies using $B = μ_{0} I /2 π a$ rather than the finite formula.
Compute: $a = 0.02 m$ , $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 15 )}{2 π ( 0.02 )} = 1.5 \times 1 0^{- 4} T$ .
Compare to Earth: ratio $= \frac{1.5 \times 1 0 ^{- 4}}{2 \times 1 0 ^{- 5}} = 7.5$ . Why? Deflection depends on the wire field relative to the background it competes with.

Verify: $B_{wire} \approx 7.5 \times$ Earth's field ✔ — clearly enough to swing a compass noticeably. Units: T throughout. Order-of-magnitude sanity: $1 0^{- 4} T$ from a 15 A wire at 2 cm is the textbook ballpark. This is why current-carrying cables disturb nearby magnetometers — an Ampère's Law shortcut gives the identical number faster for this symmetric case.

Recall Which cell is which? (self-test)

Point sitting exactly on the wire's own line ::: $B = 0$ (Cell C4, $sin θ = 0$ ). Wire from the foot to infinity, one side only ::: half the infinite value (Cell C2). Quarter arc at its centre ::: one quarter of $μ_{0} I /2 R$ (Cell C5). Radial straight leads into an arc ::: straights contribute nothing at the centre (Cell C6). Far away on a loop's axis ::: field $\propto 1/ x^{3}$ , a dipole (Cell C7).

Connections

Ampère's Law — the fast route for the symmetric cells (A, H).
Magnetic Field of a Solenoid — stack many Example-G loops.
Magnetic Dipole Moment — the $1/ x^{3}$ tail of Example G.
Right-Hand Rule — fixes every direction used above.
Lorentz Force — what these fields do to the compass needle in Example H.
Coulomb's Law — the electric $1/ r^{2}$ contrast.