Biot-Savart law — magnetic field from current element: Question bank

Before you start, pin down the vocabulary these questions reuse (all anchored to the figure below).

Definition The players in every question here

Field point $P$ — the specific location in space where we want to know the magnetic field. It is not on the wire; it sits some distance away.
Perpendicular distance $a$ — the shortest (straight-across) distance from the wire to $P$ . The point on the wire closest to $P$ is the perpendicular foot $F$ .
Current element $d l$ — one tiny segment of the wire, pointing in the direction the current flows.
$\overset{r}{^}$ — the unit vector pointing from that current element to $P$ ; and $θ$ is the angle between $d l$ and $\overset{r}{^}$ .
End-angles $θ_{1}, θ_{2}$ — for a finite straight wire, stand at $P$ , look at the perpendicular foot $F$ , then swing your line of sight to each end of the wire. The two angles you sweep out (measured from the perpendicular $P F$ ) are $θ_{1}$ and $θ_{2}$ .
Axial distance $x$ — for a circular loop, the straight-line distance measured along the axis (the line through the loop's centre, perpendicular to the loop's plane) from the centre out to the field point. At the centre $x = 0$ .

Figure — Biot-Savart law — magnetic field from current element

Three anchors from the parent, restated with these names:

The Biot–Savart magnitude is $d B = \frac{μ _{0}}{4 π} \frac{I d l sin θ}{r ^{2}}$ , where $θ$ is the angle between the segment $d l$ and $\overset{r}{^}$ .
The direction of $d B$ comes from a cross product $d l \times \overset{r}{^}$ — perpendicular to both.
Final fields ( $μ_{0} I /2 π a$ , $μ_{0} I /2 R$ ) are the result of integrating; they are not the law itself.

The finite-wire formula referenced throughout is (see figure): $B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})$ .

Definition Sign convention for the end-angles

$θ_{1}, θ_{2}$ Measure each end-angle at $P$ , starting from the perpendicular line $P F$ and swinging toward the end of the wire. Give it a sign:

If the end lies on one side of the foot $F$ (say, above), its angle is positive.
If the end lies on the opposite side of $F$ (below), its angle is also entered as positive when $F$ is between the two ends — the two contributions from opposite sides add, which is why the formula reads $sin θ_{1} + sin θ_{2}$ .
If $P$ lies beyond one end so that the foot $F$ is outside the wire (both ends on the same side of $F$ ), that nearer end's angle is measured on the same side as the other and must be entered as negative (or equivalently, use $sin θ_{2} - sin θ_{1}$ ). The two contributions then partly cancel, correctly giving a weaker field.

The safe rule: an angle is positive when its end is on the opposite side of $F$ from the reference, negative when on the same side. The angles never exceed $9 0^{\circ}$ from $P F$ for points beside the wire; when $P$ is level with an end, that angle is exactly $0$ .

True or false — justify

Doubling the current doubles the magnetic field everywhere.

True.

B

depends linearly on

I

(

d B \propto I

in the law, and integration is linear), so scaling

I

scales

B

by the same factor at every point.

The formula

B = \frac{μ _{0} I}{2 π a}

applies to a wire of any length.

False. That result assumes an infinitely long straight wire (both end-angles

θ_{1} = θ_{2} = 9 0^{\circ}

). A finite wire needs

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

, which is smaller.

A single current element

I d l

produces a field that points radially outward, like a point charge's field.

False. A point charge's field is radial, but

d B \propto d l \times \overset{r}{^}

is perpendicular to

\overset{r}{^}

— the field wraps around the element in circles, it never points along

\overset{r}{^}

Directly ahead of a current element (in the direction it points), the field it makes is zero.

True. There

θ = 0

, so

sin θ = 0

and

d l \times \overset{r}{^} = 0

. A segment sprays field sideways, never straight along its own direction.

The magnetic field of an infinite straight wire falls off as

1/ r^{2}

False. Each element contributes

1/ r^{2}

, but summing infinitely many elements along the wire gives a net

1/ a

dependence:

B = μ_{0} I /2 π a

On the axis of a circular loop, far away, the field falls as

1/ x^{2}

like Coulomb's law.

False. On-axis

B = \frac{μ _{0} I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}

(with

x

the axial distance from the centre), which for

x ≫ R

goes as

1/ x^{3}

— the signature of a magnetic dipole, not a monopole.

At the perpendicular foot's field point

P

, all the

d B

contributions from a straight wire point the same way.

True. For every element, the direction

d l \times \overset{r}{^}

(by the Right-Hand Rule) comes out identical at

P

— perpendicular to the plane containing the wire and

P

— so they add as magnitudes with no cancellation. In the figure's orientation (wire vertical,

P

to the right, current up) that common direction is straight out of the page toward you.

Biot–Savart and Ampère's Law can give different answers for the same infinite wire.

False. They must agree; both yield

μ_{0} I /2 π a

. Ampère's law is just a faster route for symmetric geometries.

At the exact centre of a current-carrying circular loop,

sin θ = 1

for every element.

True. The tangent

d l

is always perpendicular to the radius

\overset{r}{^}

on a circle, so

θ = 9 0^{\circ}

and

sin θ = 1

throughout.

The field at

P

on the perpendicular bisector of a finite wire is stronger than

μ_{0} I /2 π a

at the same distance

a

False. A finite wire has fewer contributing elements, so

sin θ_{1} + sin θ_{2} < 2

, giving a weaker field than the infinite case.

Spot the error

A student writes

d B = \frac{μ _{0}}{4 π} \frac{I d l}{r ^{2}}

and integrates for a straight wire.

The

sin θ

factor is missing. Magnetism comes from a cross product, so the directional

sin θ

is essential; without it you overcount and get the wrong constant.

Someone measures

r

from the coordinate origin to the field point

P

for every element.

r

in Biot–Savart runs from the current element itself to

P

, and it is different for each element. Using a fixed origin gives wrong

r

and wrong

θ

To find the loop's centre field, a student adds all

d B

as scalars without checking their directions.

The mathematical error is treating a vector sum as a scalar sum. It happens to be valid here only because at the centre every

d B

is parallel (all along the axis), so

∣ \sum d B ∣ = \sum ∣ d B ∣

. Off the centre — e.g. on the loop's axis at a distance — the

d B

point in different directions and this step is wrong.

A solution claims the field of a current element points along

d l

d B ⊥ d l

always, since

d l \times \overset{r}{^}

is perpendicular to

d l

. Nothing about the field lies along the segment.

For a finite wire a student uses angles measured from the wire, not from the perpendicular

P F

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

, the angles

θ_{1}, θ_{2}

are measured at

P

from the perpendicular

P F

to each end. Using the wire-axis convention flips

sin \leftrightarrow cos

and gives a wrong answer.

A student concludes

μ_{0}

has units of tesla.

μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A

— its units carry metres and amps precisely so that the whole formula comes out in tesla. Dropping them makes dimensional checks fail.

Someone treats

\overset{r}{^}

as if it always points from

P

back to the element.

\overset{r}{^}

points from the element to $P$ . Reversing it flips the sign of the cross product and reverses the field direction — you'd get the field circulating the wrong way.

A student applies

B = \frac{μ _{0} I}{4 π a} (sin θ_{1} + sin θ_{2})

with both angles positive even though

P

is beyond one end of the wire.

When the foot

F

falls outside the wire (both ends on the same side of

F

), one angle must be entered with a negative sign, so the terms partly cancel:

B = \frac{μ _{0} I}{4 π a} (sin θ_{2} - sin θ_{1})

. Keeping both positive overestimates the field.

Why questions

Why does the field circle around the wire instead of pointing toward or away from it?

The cross product

d l \times \overset{r}{^}

forces

d B

perpendicular to both the current and the line to the point. This "curl" nature reflects that there are no magnetic monopoles — only loops.

Why does a segment make maximum field to its side and none straight ahead?

Because of

sin θ

: sideways means

θ = 9 0^{\circ}

(

sin = 1

, maximum), straight ahead means

θ = 0^{\circ}

(

sin = 0

, zero). Moving charge throws field perpendicular to its motion.

Why does the infinite wire's

1/ r^{2}

per element become an overall

1/ a

First get the geometry. Put the foot

F

at the origin and let a segment sit at height

y

up the wire; the angle

θ

is measured at that segment between

d l

(up the wire) and

\overset{r}{^}

(toward

P

). In the right triangle

F

–segment–

P

, the side

a = P F

is opposite

θ

and

r

is the hypotenuse from segment to

P

, so

sin θ = a / r

, i.e.

r = a / sin θ

. Also the height along the wire is

y = a cot θ

(adjacent over opposite

= y / a

gives

cot θ

, wait —

y

is measured from

F

, and

tan (angle at P) = y / a

; equivalently

y = a cot θ

using the segment-angle

θ

). Differentiating

y = a cot θ

gives

d y = - a d θ / sin^{2} θ

, so

d l = ∣ d y ∣ = a d θ / sin^{2} θ

. Substitute

r = a / sin θ

and

d l = a d θ / sin^{2} θ

into

d B = \frac{μ _{0} I}{4 π} \frac{s i n θ d l}{r ^{2}}

: the

a

's and

sin

's collapse to

d B = \frac{μ _{0} I}{4 π a} sin θ d θ

. Integrating

θ

from

0

π

gives

\int_{0}^{π} sin θ d θ = 2

, so

B = \frac{μ _{0} I}{2 π a}

— a single power of

a

survives because one factor of

a

from

d l

cancels one of the two in

1/ r^{2}

Why can Ampère's law replace this messy integral for a straight wire but not for a finite one?

Ampère's law needs high symmetry so

B

is constant along a chosen loop. An infinite wire has that symmetry; a finite wire does not, so you must fall back on direct Biot–Savart integration.

Why do only the axial components of

d B

survive on a loop's axis?

By symmetry, for every element there is a diametrically opposite element whose

d B

has an equal-and-opposite radial (off-axis) component; those cancel in pairs, leaving only the components pointing along the axis, which all add.

Why is Biot–Savart called the magnetic analogue of Coulomb's Law, yet fundamentally different?

Both share the

1/ r^{2}

falloff and a source element. But Coulomb's field points along

\overset{r}{^}

(a source point), while Biot–Savart's cross product makes the field circulate — magnetism has no isolated poles.

Edge cases

What is the field at a point lying on the wire's own line (the longitudinal line through the wire, beyond its end)?

For every element,

\overset{r}{^}

is parallel or antiparallel to

d l

, so

θ = 0

18 0^{\circ}

and

sin θ = 0

. The net field on the wire's own line is zero.

What happens to

B = μ_{0} I /2 π a

a \to 0

(approaching the wire)?

It diverges to infinity. This is an idealization: a real wire has finite thickness, so the current is spread out and the field stays finite inside it.

What is the field at the centre of a half-loop (semicircle) of radius

R

Half the full-loop value:

B = \frac{μ _{0} I}{4 R}

. Only half the arc-length

π R

contributes, and every element still has

sin θ = 1

at the centre.

Do the two straight lead-in wires that connect to a loop at its centre contribute to the centre field?

No — if they run radially toward the centre,

d l

is parallel to

\overset{r}{^}

there (

θ = 0

), so those straight sections add nothing at the centre.

What does the on-axis loop formula give at the very centre (

x = 0

B = \frac{μ _{0} I R ^{2}}{2 R ^{3}} = \frac{μ _{0} I}{2 R}

, exactly matching the direct centre result — a built-in consistency check.

In the finite-wire formula, what does one angle equalling

0

mean physically?

It means the point

P

is level with one end of the wire (the perpendicular foot

F

coincides with that end). Then only the other end's

sin θ

contributes.

What if

P

lies beyond one end of the wire, so the foot

F

falls off the wire entirely?

Both ends now sit on the same side of

F

. Enter the nearer end's angle as negative, giving

B = \frac{μ _{0} I}{4 π a} (sin θ_{2} - sin θ_{1})

— the contributions partly cancel and the field is weaker than for a symmetric placement.

If a wire carries zero current, what field does Biot–Savart predict?

Exactly zero everywhere, since

d B \propto I

. No current means no source element and no field — the law degenerates cleanly.

Connections

Parent topic — full derivations these traps test.
Ampère's Law — the symmetry shortcut that must agree with Biot–Savart.
Magnetic Field of a Solenoid — integrated loops; test your loop reasoning there.
Magnetic Dipole Moment — why the far-axis loop field goes as $1/ x^{3}$ .
Coulomb's Law — the $1/ r^{2}$ analogue to contrast against.
Lorentz Force — what the resulting $B$ does.
Right-Hand Rule — fixes every direction above.