PhysicsISC Class 12 · Moving Charges and Magnetism

Biot-Savart Law: Field at the Centre and on the Axis of a Circular Coil

Updated 15 October 2026 · Fact-checked

The Biot-Savart law gives the field of a small current element: $d\vec{B} = \frac{\mu_0}{4\pi}\frac{I\,d\vec{l} \times \hat{r}}{r^2}$. Adding the contributions round a circular coil of N turns and radius R gives $B = \frac{\mu_0NI}{2R}$ at the centre and $B = \frac{\mu_0NIR^2}{2(R^2 + x^2)^{3/2}}$ on the axis, along the axis.

Understand Biot-Savart Law and the Field of a Circular Coil

The law. A current element $I d l$ produces at a point P, at position $r$ from it, a field
$d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}, d B = \frac{μ _{0}}{4 π} \frac{I d l s i n θ}{r ^{2}},$
where θ is the angle between $d l$ and $r$ and $\frac{μ _{0}}{4 π} = 1 0^{- 7}$ T m A⁻¹. The field is perpendicular to both $d l$ and $r$ (cross product), falls as $\frac{1}{r ^{2}}$ , and is zero along the line of the element (θ = 0). It is the magnetic counterpart of Coulomb's law, but the source is a vector (the current element), not a scalar charge.

Direction. Use the right-hand rule: curl the fingers of the right hand in the direction of the current in a loop; the thumb gives the field at the centre. For a straight wire, point the thumb along the current; the fingers curl in the direction of the field lines (circles round the wire). Into the page is written ⊗, out of the page ⊙; the 2024 examiners wanted "perpendicular to the plane of the paper, inwards", not just "inward".

Field at the centre of a circular coil (derivation). Take a coil of radius R carrying current I. For every element $d l$ , the line to the centre O is a radius, so $d l ⊥ r$ (θ = 90°) and r = R:
$d B = \frac{μ _{0}}{4 π} \frac{I d l s i n 90°}{R ^{2}} .$
Every element gives a field in the same direction (perpendicular to the plane of the coil), so the magnitudes add:
$B = \frac{μ _{0} I}{4 π R ^{2}} \int_{0}^{2 π R} d l = \frac{μ _{0} I}{4 π R ^{2}} (2 π R) = \frac{μ _{0} I}{2 R} .$
For N turns, $B = \frac{μ _{0} N I}{2 R}$ . (Show θ = 90°, the limits 0 to 2πR, and the direction; the 2024 examiners deducted marks for missing these.)

Field of an arc. An arc subtending angle φ (radians) at the centre contributes $B = \frac{μ _{0} I}{4 π R} φ$ : a semicircle gives $\frac{μ _{0} I}{4 R}$ .

Field on the axis (derivation outline). At P on the axis, distance x from the centre, every element is at distance $r = R^{2} + x^{2}$ and $d l ⊥ r$ , so $d B = \frac{μ _{0}}{4 π} \frac{I d l}{R ^{2} + x ^{2}}$ . Resolve $d B$ along the axis and perpendicular to it. By symmetry, the perpendicular components from diametrically opposite elements cancel. The axial components add, each being $d B sin α$ with $sin α = \frac{R}{R ^{2} + x ^{2}}$ :
$B = \frac{μ _{0} I}{4 π ( R ^{2} + x ^{2} )} \cdot \frac{R}{R ^{2} + x ^{2}} \cdot 2 π R = \frac{μ _{0} N I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$ (N turns).
At x = 0 this gives the centre value. Far away ( $x ≫ R$ ), $B = \frac{μ _{0}}{4 π} \frac{2 N I A}{x ^{3}} = \frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}}$ , with $A = π R^{2}$ and $m = N I A$ : the same form as the axial field of an electric dipole. A current loop is a magnetic dipole with moment $m = N I A$ (A m²), its direction given by the right-hand rule.

Long straight wire. From Ampère's circuital law (or Biot-Savart), $B = \frac{μ _{0} I}{2 π r}$ at distance r, with field lines as concentric circles. For a wire of finite length (formula only), $B = \frac{μ _{0} I}{4 π r} (sin φ_{1} + sin φ_{2})$ . Do not confuse this with the coil formula $\frac{μ _{0} N I}{2 R}$ ; the 2025 examiners saw that confusion.

Several sources. Fields add as vectors. Between two long parallel wires carrying currents in the same direction the fields oppose; with opposite currents they add. Find each magnitude, mark each direction (⊗ or ⊙), then combine. The 2025 examiners found many candidates computed magnitudes but ignored directions.

Biot-Savart Law and the Field of a Circular Coil: formulas and key facts

Biot-Savart law: $d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$; $\frac{μ _{0}}{4 π} = 1 0^{- 7}$ T m A⁻¹.
Centre of a coil: $B = \frac{μ _{0} N I}{2 R}$; Perpendicular to the plane of the coil.
Axis of a coil: $B = \frac{μ _{0} N I R ^{2}}{2 ( R ^{2} + x ^{2} ) ^{3/2}}$; Along the axis.
Arc at its centre: $B = \frac{μ _{0} I φ}{4 π R}$; φ in radians; semicircle $\frac{μ _{0} I}{4 R}$ .
Long straight wire: $B = \frac{μ _{0} I}{2 π r}$; Circular field lines.
Finite wire (formula only): $B = \frac{μ _{0} I}{4 π r} (sin φ_{1} + sin φ_{2})$; Angles measured from the perpendicular.
Magnetic moment of a loop: $m = N I A$; Far axial field $\frac{μ _{0}}{4 π} \frac{2 m}{x ^{3}}$ .

How to solve Biot-Savart Law and the Field of a Circular Coil questions

Identify the source shape, choose the formula, then handle direction separately.

1Decide what produces the field: a full coil, an arc, a long straight wire, or a combination.
2Write the matching formula and substitute in SI units ( $μ_{0} = 4 π \times 1 0^{- 7}$ T m A⁻¹).
3Find the direction of each field with the right-hand rule; mark ⊗ or ⊙ (or the axis direction).
4Add fields as vectors: same direction add, opposite directions subtract.
5For "what if I, N or R changes" questions, write the proportionality first ( $B \propto \frac{N I}{R}$ ).
6In derivations, draw the element $I d l$ , the vector $r$ , the angle and the resulting $d B$ .

Biot-Savart Law and the Field of a Circular Coil in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Field of a long straight wire

Suppose a long straight cable under a laboratory bench carries 20 A northwards, and a compass rests on the bench 4 cm directly above it. Suppose the horizontal component of the Earth's field at the place is $4 \times 1 0^{- 5}$ T.

Find the field of the cable at the compass and the angle through which the needle turns from north.

Try it first, then show the solutionSolution

$B = \frac{μ _{0} I}{2 π r} = \frac{2 \times 1 0 ^{- 7} \times 20}{0.04} = 1.0 \times 1 0^{- 4}$ T, horizontal and perpendicular to the cable.
The needle sets along the resultant: $tan θ = \frac{1.0 \times 1 0 ^{- 4}}{4 \times 1 0 ^{- 5}} = 2.5$ , so θ ≈ 68°.

Answer: $1.0 \times 1 0^{- 4}$ T; the needle turns about 68° from north.

Memory hook: Straight wire: $2 \times 1 0^{- 7} \frac{I}{r}$ .

Designing a coil

Imagine a student who wants a field of 1 mT at the centre of a 100-turn coil of radius 8 cm for an experiment.

What current is needed?

Try it first, then show the solutionSolution

$I = \frac{2 R B}{μ _{0} N} = \frac{2 \times 0.08 \times 1 0 ^{- 3}}{4 π \times 1 0 ^{- 7} \times 100}$ .
$I = \frac{1.6 \times 1 0 ^{- 4}}{1.257 \times 1 0 ^{- 4}} \approx 1.27$ A.

Answer: About 1.27 A.

Memory hook: Coil centre: $\frac{μ _{0} N I}{2 R}$ .

Tricks to remember Biot-Savart Law and the Field of a Circular Coil

Short memory aids. Say each one aloud once, then use it on a question the same day.

Coil: 2R; wire: 2πr
Coil centre $\frac{μ _{0} N I}{2 R}$ ; long wire $\frac{μ _{0} I}{2 π r}$ . The π is in the wire.
Use it when: Choosing the right formula.
Curl and thumb
Loop: fingers with current, thumb = field. Wire: thumb with current, fingers = field.
Use it when: Directions.
Axis falls as the cube
Far along the axis B ∝ $\frac{1}{x ^{3}}$ : a loop is a dipole.
Use it when: Far-field and ratio questions.
Same currents attract, fields cancel between
Parallel wires with the same current direction: fields oppose in the gap.
Use it when: Two-wire problems.

Biot-Savart Law and the Field of a Circular Coil: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Direction at the centre · 1 mark · MCQBoard level

A circular coil lies in the plane of the paper and carries current anticlockwise as seen by the reader. The field at its centre is:
(a) in the plane, to the right
(b) perpendicular to the paper, out of it (towards the reader)
(c) perpendicular to the paper, into it
(d) zero

Try it first, then show the solutionSolution

Curl the right-hand fingers anticlockwise: the thumb points out of the paper.

Answer: (b).

Example 2: Field at the centre · 2 marks · Short answerBoard level

A coil of 100 turns and radius 5 cm carries 0.5 A. Find the magnetic field at its centre.

Try it first, then show the solutionSolution

$B = \frac{μ _{0} N I}{2 R} = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 0.5}{2 \times 0.05}$ . (1)
$B = 2 π \times 1 0^{- 4} \approx 6.28 \times 1 0^{- 4}$ T. (1)

Answer: $6.28 \times 1 0^{- 4}$ T, perpendicular to the plane of the coil.

Example 3: Field of an arc · 3 marks · Short answerHarder board level

Using the Biot-Savart law, find the field at the centre of a circular arc of radius R that subtends an angle φ at the centre and carries current I. Hence find the field at the centre of a semicircular wire of radius 5 cm carrying 4 A (the straight leads point radially and contribute nothing).

Try it first, then show the solutionSolution

Each element is perpendicular to the radius: $d B = \frac{μ _{0}}{4 π} \frac{I d l}{R ^{2}}$ , all in the same direction. (1)
Arc length $R φ$ : $B = \frac{μ _{0} I}{4 π R ^{2}} (R φ) = \frac{μ _{0} I φ}{4 π R}$ . (1)
Semicircle (φ = π): $B = \frac{μ _{0} I}{4 R} = \frac{4 π \times 1 0 ^{- 7} \times 4}{4 \times 0.05} \approx 2.51 \times 1 0^{- 5}$ T. Radial leads lie along $r$ , so $d l \times \overset{r}{^} = 0$ for them. (1)

Answer: $\frac{μ _{0} I φ}{4 π R}$ ; about $2.5 \times 1 0^{- 5}$ T.

Example 4: Two long wires with opposite currents · 3 marks · Short answerHarder board level

Two long parallel wires 12 cm apart carry 6 A and 3 A in opposite directions. Find the field at a point between them, 4 cm from the 6 A wire.

Try it first, then show the solutionSolution

$B_{1} = \frac{2 \times 1 0 ^{- 7} \times 6}{0.04} = 3.0 \times 1 0^{- 5}$ T; $B_{2} = \frac{2 \times 1 0 ^{- 7} \times 3}{0.08} = 7.5 \times 1 0^{- 6}$ T. (1)
With opposite currents, the right-hand rule gives both fields in the same direction between the wires. (1)
$B = 3.75 \times 1 0^{- 5}$ T, perpendicular to the plane containing the wires. (1)

Answer: $3.75 \times 1 0^{- 5}$ T.

Example 5: Where the axial field is one-eighth · 2 marks · Short answerHOTS / competency

At what distance from the centre, on the axis of a circular coil of radius R, is the field one-eighth of its value at the centre?

Try it first, then show the solutionSolution

$\frac{B _{x}}{B _{0}} = \frac{R ^{3}}{( R ^{2} + x ^{2} ) ^{3/2}} = \frac{1}{8}$ , so $(R^{2} + x^{2})^{3/2} = 8 R^{3}$ . (1)
$R^{2} + x^{2} = 4 R^{2}$ , so $x = 3 R$ . (1)

Answer: $x = 3 R$ .

Example 6: Cancelling a coil's field · 4 marks · Case-basedHOTS / competency

Suppose a single circular loop of radius 10 cm carries 1 A, and a long straight wire lies in the same plane at 20 cm from the loop's centre.
(i) Find the loop's field at its centre. (1 mark)
(ii) Write the wire's field at the centre in terms of its current $I_{w}$ . (1 mark)
(iii) Find $I_{w}$ that makes the net field at the centre zero, and say how its direction must be related to the loop current. (1 mark)
(iv) Find the magnetic moment of the loop. (1 mark)

Try it first, then show the solutionSolution

(i) $B_{L} = \frac{μ _{0} I}{2 R} = \frac{4 π \times 1 0 ^{- 7}}{0.2} = 2 π \times 1 0^{- 6} \approx 6.28 \times 1 0^{- 6}$ T.
(ii) $B_{W} = \frac{μ _{0} I _{w}}{2 π ( 0.2 )} = 1.0 \times 1 0^{- 6} I_{w}$ T.
(iii) $I_{w} = 2 π \approx 6.28$ A, flowing so that its field at the centre is opposite to the loop's (for example, along the side of the wire nearest the loop, the wire current must run opposite to the loop current on its near side).
(iv) $m = I A = 1 \times π (0.1)^{2} \approx 3.14 \times 1 0^{- 2}$ A m².

Answer: (i) $6.28 \times 1 0^{- 6}$ T; (ii) $1 0^{- 6} I_{w}$ ; (iii) about 6.28 A, opposing the loop field; (iv) $3.14 \times 1 0^{- 2}$ A m².

Where students lose marks in Biot-Savart Law and the Field of a Circular Coil

The slips examiners see most often, and what to write instead.

Mistake: Incomplete derivation diagram
The ISC 2024 examiners reported missing θ = 90°, unmarked $d l$ , no direction of B, and no limits of integration.
Do this: Draw the coil, mark $I d l$ , $r$ , θ = 90° and the direction of B; integrate from 0 to 2πR.
Usually costs: 1 to 2 marks.
Mistake: Ampère instead of Biot-Savart
Also in 2024: some candidates used Ampère's circuital law or derived the axial field and simplified, when asked to use Biot-Savart at the centre.
Do this: Use the law named in the question, with the symbols given (R, not r or a).
Usually costs: 1 mark.
Mistake: Wrong formula for a coil
The 2025 examiners saw the straight-wire formula used for a circular coil.
Wrong working
$B = \frac{μ _{0} N I}{2 π R}$
Right working
$B = \frac{μ _{0} N I}{2 R}$
Do this: Learn both formulas side by side.
Usually costs: 1 to 2 marks.
Mistake: Ignoring direction
In 2025 candidates added magnitudes of fields from two wires without considering direction, missing that they cancelled.
Do this: Mark ⊗ or ⊙ for each field before combining.
Usually costs: 1 to 2 marks.

Exam tips

State the direction as "perpendicular to the plane of the paper, inwards/outwards".
Use $\frac{μ _{0}}{4 π} = 1 0^{- 7}$ to speed up arithmetic.
Keep the symbols used in the question (I′, R) in your final expression.
For ratio questions, write the proportionality and avoid substituting numbers.

Practice questions on Biot-Savart Law and the Field of a Circular Coil

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
According to the Biot-Savart law, the field of a current element at a distance r varies as:
(a) r
(b) $\frac{1}{r}$
(c) $\frac{1}{r ^{2}}$
(d) $\frac{1}{r ^{3}}$
Hint
Like Coulomb's law.
Answer
(c).
Q2 · 2 marks · Short answerBoard level
The field at the centre of a circular coil is B. What is the field at the same point when the current is tripled, the number of turns halved and the radius increased 1.5 times? (A variation on ISC 2025.)
Hint
$B = \frac{μ _{0} N I}{2 R}$ .
Answer
$\frac{3 \times \frac{1}{2}}{1.5} B = B$ : unchanged.
Q3 · 1 mark · Very short answerHarder board level
At what distance from a long straight wire carrying 25 A is the field $1 \times 1 0^{- 4}$ T?
Hint
$r = \frac{μ _{0} I}{2 π B}$ .
Answer
5 cm.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The field at the centre of a current-carrying circular coil is perpendicular to the plane of the coil.
Reason (R): Every current element of the coil is perpendicular to the radius joining it to the centre, so each $d B$ there is perpendicular to the plane, in the same sense.
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
$d l \times \overset{r}{^}$ .
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Find the ratio of the field on the axis of a coil, at a distance equal to its radius, to the field at its centre.
Hint
Put x = R.
Answer
$\frac{R ^{3}}{( 2 R ^{2} ) ^{3/2}} = \frac{1}{2 2} \approx 0.354$ .
Q6 · 2 marks · Short answerHOTS / competency
Two concentric, coplanar circular coils of radii 10 cm and 20 cm carry 2 A and I. Find I (and its sense) for the field at the centre to be zero.
Hint
$\frac{I _{1}}{R _{1}} = \frac{I _{2}}{R _{2}}$ .
Answer
I = 4 A, flowing in the opposite sense to the 2 A current.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose two long parallel wires P and Q, 10 cm apart, carry 4 A and 6 A in opposite directions.
(i) Find the field at the midpoint between them. (1 mark)
(ii) Explain why there is no zero-field point between the wires. (1 mark)
(iii) Find the zero-field point on the line through both wires. (1 mark)
(iv) Find the force per metre between the wires and say whether it is attractive or repulsive. (1 mark)
Hint
Opposite currents: fields add between the wires.
Answer
(i) $2 \times 1 0^{- 7} \frac{4 + 6}{0.05} = 4 \times 1 0^{- 5}$ T. (ii) Between the wires both fields point the same way. (iii) Outside, beyond P: $\frac{4}{x} = \frac{6}{x + 0.1}$ , x = 20 cm from P. (iv) $\frac{2 \times 1 0 ^{- 7} \times 24}{0.1} = 4.8 \times 1 0^{- 5}$ N m⁻¹, repulsive.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the Biot-Savart law in vector form.Show answer
$d B = \frac{μ _{0}}{4 π} \frac{I d l \times r ^}{r ^{2}}$ .
2.Field at the centre of a coil of N turns?Show answer
$\frac{μ _{0} N I}{2 R}$ .
3.Field 10 cm from a long wire carrying 5 A?Show answer
$1 \times 1 0^{- 5}$ T.
4.Magnetic moment of a coil?Show answer
$m = N I A$ .
5.Field at the centre of a semicircular wire?Show answer
$\frac{μ _{0} I}{4 R}$ .

Biot-Savart Law and the Field of a Circular Coil: frequently asked questions

Why is a current loop called a magnetic dipole?

Far along its axis its field is \frac{\mu_0}{4\pi}\frac{2m}{x^3}, exactly like the axial field of a bar magnet or an electric dipole, and in a uniform field it feels a torque \vec{m} \times \vec{B}.

Why does a straight current element produce no field along its own line?

Along the line, d\vec{l} and \vec{r} are parallel, so d\vec{l} \times \hat{r} = 0.

Is the field at the centre of a coil uniform around the coil?

No. The field is strongest near the wire and varies from point to point; only near the centre is it roughly uniform. (A 2024 assertion-reason item turned on this.)

Which derivations are in the ISC 2027 syllabus?

The field at the centre of a circular loop and at any point on its axis using the Biot-Savart law. The finite straight wire is formula only.

How was this tested recently?

In 2024 a 3-mark derivation of B at the centre of a coil; in 2025 a 2-mark question on changing I, N and R, and a 3-mark net field of two wires.