Magnetic field on axis of circular loop

Take a single loop of wire, connect it to a battery, and something invisible happens along the line running straight through its centre, perpendicular to the loop's plane. A magnetic field springs up all along that line — strongest right at the centre, weakening steadily as you move outward, and eventually behaving exactly like the field of a tiny bar magnet planted at the loop's heart. This is not a separate law of nature; it is the Biot-Savart law, applied patiently around the full circumference of a circle, with the sideways contributions from opposite sides of the loop cancelling in pairs and only the along-the-axis pieces surviving to add up. Why does a CBSE Class 12 student need this result cold? Because it is the bridge between two ideas the chapter builds toward. On one side sits the raw Biot-Savart law for a tiny current element — useful, but awkward to apply point by point in an exam. On the other side sits the picture of a current loop as a magnetic dipole, with a magnetic moment $m = IA$ , which later explains torque on a coil and the working of a moving-coil galvanometer. The axial-field formula for a circular loop is the one calculation that connects the two: it is exact, it is examinable by name, and its far-away limit is precisely the dipole field you meet again in the very next topic of the chapter. This section keeps the two worked numbers — a $50$ -turn coil giving about $0.628\ \text{mT}$ at its centre, and a single loop giving about $3.57\ \mu\text{T}$ at a point $12\ \text{cm}$ off-centre on its axis — and builds outward from them: a full step-by-step derivation of the formula, a sharper look at the geometry that makes the perpendicular components cancel, a second worked pair of examples pushed to a competitive-exam pace, and the traps that quietly cost marks on this exact topic every year.

Key takeaways

The magnetic field on the axis of a circular loop of radius $R$ carrying current $I$ , at distance $x$ from the centre, is $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ .
At the centre of the loop ( $x = 0$ ) this collapses to the simpler, very frequently quoted $B = \dfrac{\mu_0 I}{2R}$ .
The field only ever points along the axis — never sideways — because the perpendicular components from diametrically opposite current elements cancel exactly by symmetry.
The direction is fixed by the right-hand curl rule: curl the fingers along the current's circulation, and the thumb gives the direction of $\vec B$ on the axis.
Far from the loop, when $x \gg R$ , the field falls off as $B \approx \dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}$ with $m = I\pi R^2$ — the loop behaves exactly like a magnetic dipole seen from a distance.
For an $N$ -turn coil, every axial-field result simply scales by a factor of $N$ , since the $N$ loops carry the same current and stack their fields.
Key idea: integrating the Biot-Savart law around a circular loop, with perpendicular components cancelling by symmetry, gives an exact axial field $B = \mu_0 I R^2 / [2(R^2+x^2)^{3/2}]$ that reduces to $\mu_0 I / 2R$ at the centre and to a $1/x^3$ dipole field far away.

The core insight

The axial field of a circular current loop, $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ , comes from integrating the Biot-Savart law around the loop with only the along-axis components of each element's contribution surviving, and it smoothly connects the centre-field result $\mu_0 I/2R$ to the loop's far-field identity as a magnetic dipole.

Where it fits in the syllabus

This is a directly prescribed, named result in the CBSE Class 12 unit Moving Charges and Magnetism, listed explicitly alongside the Biot-Savart law itself. Board papers expect a student to state the formula, use it in numerical problems (including the centre special case and multi-turn coils), and connect it conceptually to the loop's identity as a magnetic dipole. A full re-derivation from first principles is sometimes asked as a 3-5 mark derivation question, so both the formula and its origin matter. The off-axis field, the toroid, and a full Ampere's-law-style proof of Biot-Savart are explicitly outside this chapter's prescribed scope and are not required here.

The physical picture

Picture a circular loop of wire of radius $R$ , lying flat, carrying a steady current $I$ . Every tiny arc of that wire is a current element that, by the Biot-Savart law, contributes a small magnetic field at any nearby point. At a point P on the axis of the loop, a distance $x$ from the centre, every single current element around the loop is exactly the same perpendicular distance from P — the slant distance $\sqrt{R^2+x^2}$ — and each contributes a small field $dB$ perpendicular to that slant line. What differs from element to element is only the direction of that small contribution within a cone swept around the axis. Two current elements on opposite sides of the loop contribute equal-magnitude fields whose components perpendicular to the axis point in opposite directions and cancel, while their components along the axis point the same way and add. Summing (integrating) these along-axis components around the full circle gives the net field:

B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}

where: $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ is the permeability of free space, $I$ is the steady current in the loop, $R$ is the loop's radius, and $x$ is the distance of the point P from the centre, measured along the axis. The field $B$ points along the axis, with its sense fixed by the right-hand curl rule applied to the current's circulation direction. Two special cases matter enormously for exam use. At the centre of the loop, $x = 0$ , and the formula reduces to

B = \dfrac{\mu_0 I}{2R}

where: this is simply the general formula with $x$ dropped, and it is the single most commonly quoted result of this topic — worth memorising directly rather than re-deriving every time. Far from the loop, when $x \gg R$ , the $R^2$ term in the denominator becomes negligible compared with $x^2$ , and the formula simplifies to

B \approx \dfrac{\mu_0 I R^2}{2x^3} = \dfrac{\mu_0}{4\pi}\dfrac{2m}{x^3}

where: $m = I\pi R^2$ is the magnetic dipole moment of the loop (current times the enclosed area). This is exactly the mathematical form of the axial field of a bar magnet's equivalent dipole — the reason a current loop is called a magnetic dipole, a connection explored fully in the next topic of the chapter, "current loop as a magnetic dipole." Between the centre and the far zone the field decreases smoothly and monotonically; it never rises again as $x$ increases, and it is never sideways.

Figure

A circular current loop lying flat with an axis drawn through its centre and field vectors marked at the centre, a nearby point and a distant point on the axis, showing the field strongest and pointing along the axis at the centre and shrinking steadily further out

Circular loop axial field

Derivation AMagnetic field on the axis of a circular current loop

The goal here is to derive, from the Biot-Savart law alone, the exact expression for the magnetic field at a point P on the axis of a circular loop of radius $R$ carrying current $I$ , at a distance $x$ from the centre. The central idea a student must grip firmly is a symmetry argument: every current element around the loop lies the same slant distance from P, so it contributes an equal-magnitude $dB$ , but these small contributions point in different directions around a cone, and only their components along the axis survive after adding all of them up — the sideways parts cancel between diametrically opposite elements. Getting that cancellation right, and expressing the surviving axial component with the correct trigonometric factor, is the entire content of the derivation; the final integral is then a trivial one because every element's geometric factor is identical.

Step 1. Set up the geometry. Let the loop lie in a plane, centred at the origin, and let P be a point on the axis (perpendicular to the loop's plane through its centre) at distance $x$ from the centre. Consider a small current element $I\,d\vec l$ at any point on the loop. The straight-line distance from this element to P is the same for every element on the loop, namely $r = \sqrt{R^2+x^2}$ , because P lies directly on the axis and every point of the loop is at radius $R$ from the centre.

Step 2. Apply the Biot-Savart law to this element. The magnitude of its contribution to the field at P is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{r^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2+x^2}$ , since the current element $d\vec l$ is tangent to the circle and always perpendicular to the line joining it to P, so $\sin\theta = 1$ in the Biot-Savart formula and this simplification is exact, not approximate.

Step 3. Resolve $d\vec B$ into components. This small vector $d\vec B$ is perpendicular to the line joining the element to P, and by the geometry of the setup it makes an angle with the axis such that its component along the axis is $dB\cos\phi$ and its component perpendicular to the axis is $dB\sin\phi$ , where $\phi$ is the angle between the line from the element to P and the axis itself. From the right triangle formed by $R$ , $x$ and $r$ , $\cos\phi = R/\sqrt{R^2+x^2}$ .

Step 4. Cancel the perpendicular components. Take a diametrically opposite current element on the loop. Its contribution $d\vec B'$ has the same magnitude (by the symmetry of Step 1) and the same axial component, but its perpendicular component points in exactly the opposite direction — sideways, away from the axis, on the opposite side. Pairing up every element with its diametric partner this way, all the perpendicular components cancel in pairs around the full loop, leaving only the axial components to add.

Step 5. Integrate the axial components around the loop. The axial component of each element is $dB\cos\phi = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{R^2+x^2}\times\dfrac{R}{\sqrt{R^2+x^2}}$ . Since this factor is identical for every element, integrating around the full loop just replaces $dl$ with the total circumference $2\pi R$ : $B = \dfrac{\mu_0}{4\pi}\dfrac{I(2\pi R)}{R^2+x^2}\times\dfrac{R}{\sqrt{R^2+x^2}} = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ .

Step 6. Check the limits. Setting $x = 0$ gives $B = \dfrac{\mu_0 I R^2}{2R^3} = \dfrac{\mu_0 I}{2R}$ , the correct centre-field result. For $x \gg R$ , the denominator $(R^2+x^2)^{3/2} \approx x^3$ , giving $B \approx \dfrac{\mu_0 I R^2}{2x^3}$ , the dipole-like far-field form. Both limits match the results already known independently, confirming the derivation.

Worked examples

Board level - centre field of a 50-turn coil

A circular coil of radius $R = 0.1\ \text{m}$ has $N = 50$ turns and carries a current of $I = 2\ \text{A}$ . Find the magnetic field at the centre of the coil.

Step 1. Use the centre formula for $N$ turns: each of the $N$ loops carries the same current and contributes the same centre field, so the total is simply $N$ times a single loop's value. $B = \dfrac{\mu_0 N I}{2R}$ .

Step 2. Substitute values, using $\mu_0 = 4\pi\times10^{-7}\ \text{T}\,\cdot\,\text{m/A}$ . $B = \dfrac{(4\pi\times10^{-7})(50)(2)}{2(0.1)}$ .

Step 3. Compute the numerator. $4\pi\times10^{-7}\times50\times2 = 4\pi\times10^{-7}\times100 = 4\pi\times10^{-5}$ .

Step 4. Divide by the denominator ( $0.2$ ) and evaluate with $\pi \approx 3.1416$ . $B = \dfrac{4\pi\times10^{-5}}{0.2} = 20\pi\times10^{-5} \approx 6.28\times10^{-4}\ \text{T}$ .

The field at the centre is about $6.28\times10^{-4}\ \text{T}$ , or $0.628\ \text{mT}$ .

Competitive level - axial field off-centre, then compare with the centre value

A single-turn circular loop of radius $R = 0.05\ \text{m}$ carries a current of $I = 5\ \text{A}$ . Find the magnetic field at a point on the axis at a distance $x = 0.12\ \text{m}$ from the centre, and state what fraction this is of the field the same loop would produce at its own centre.

Step 1. Use the general axial formula. $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ .

Step 2. Compute $R^2$ and $x^2$ . $R^2 = (0.05)^2 = 0.0025$ , $x^2 = (0.12)^2 = 0.0144$ , so $R^2+x^2 = 0.0169$ .

Step 3. Compute $(R^2+x^2)^{3/2}$ . First $\sqrt{0.0169} = 0.13$ ; then $(0.0169)^{3/2} = 0.0169\times0.13 = 0.002197$ .

Step 4. Substitute into the formula. $B = \dfrac{(4\pi\times10^{-7})(5)(0.0025)}{2(0.002197)}$ . Numerator: $4\pi\times10^{-7}\times5\times0.0025 = 4\pi\times1.25\times10^{-9} = 1.5708\times10^{-8}$ . Denominator: $2\times0.002197 = 0.004394$ . $B = \dfrac{1.5708\times10^{-8}}{0.004394} \approx 3.57\times10^{-6}\ \text{T}$ .

Step 5. Compare with the centre value. At the centre of this same loop, $B_0 = \dfrac{\mu_0 I}{2R} = \dfrac{(4\pi\times10^{-7})(5)}{2(0.05)} = \dfrac{6.283\times10^{-6}}{0.1} = 6.28\times10^{-5}\ \text{T}$ . The ratio is $\dfrac{3.57\times10^{-6}}{6.28\times10^{-5}} \approx 0.057$ , so the field at $x = 0.12\ \text{m}$ is only about $5.7\%$ of the centre field, even though $x$ is just $2.4$ times $R$ — a sharp illustration of how quickly the axial field falls off.

The axial field at $x = 0.12\ \text{m}$ is about $3.57\times10^{-6}\ \text{T}$ , roughly $5.7\%$ of the $6.28\times10^{-5}\ \text{T}$ this loop produces at its own centre. Timing note: this two-part comparison should take under three minutes in an exam. Compute $R^2+x^2$ first and keep it as a single number; do not expand $(R^2+x^2)^{3/2}$ by any other route than square-root-then-cube, since that is the fastest arithmetic path. Reusing the already-known centre-field value for the ratio, rather than recomputing everything from scratch, is the time-saving habit worth building.

Common traps

Using the centre-of-loop formula $B = \mu_0 I/(2R)$ for a point that is not at the centre — the general axial formula with $(R^2+x^2)^{3/2}$ must be used whenever $x \neq 0$ , and mixing these two up is the single most common error on this topic.
Forgetting to multiply by the number of turns $N$ when the coil has multiple loops stacked together — a coil is not a single loop, and every turn adds its own field.
Mixing up $R^2 + x^2$ with $(R+x)^2$ in the denominator — these are very different numbers and a common algebra slip under exam pressure.
Assuming the field is uniform along the axis — it actually decreases as $x$ increases, is maximum at the centre, and eventually falls off as $1/x^3$ far from the loop, as the worked comparison above shows numerically.
Ignoring the direction of the field entirely — the axial field is always along the axis, never sideways, but its sense (which way along the axis it points) still depends on the current's circulation direction via the right-hand rule.
Forgetting that the formula is valid only on the axis — trying to apply it directly to an off-axis point (say, in the plane of the loop but away from the centre) gives a wrong answer, since that calculation needs integration this syllabus does not require.

Common trapThe axial formula only works on the axis, and the centre formula is a special case of it, not a separate rule

Students frequently memorise $B = \mu_0 I/(2R)$ and $B = \mu_0 I R^2/[2(R^2+x^2)^{3/2}]$ as two unrelated formulas to be picked between by guesswork. They are not independent: the centre formula is exactly what the general axial formula gives when $x = 0$ , nothing more. The safest exam habit is to always start from the general formula, substitute $x = 0$ only when the question genuinely asks for the centre, and never apply either formula to a point that is off the axis — the field off-axis needs a different, more advanced calculation that is outside the CBSE syllabus, and no shortcut using these two formulas gives the right answer there.

You should be able to…

State and apply the general formula $B = \dfrac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$ for the axial field of a circular current loop, including for a numerical distance $x$ .
Reduce this formula correctly to $B = \mu_0 I/(2R)$ at the centre, and recognise it as a special case, not a separate law.
Extend the single-loop formula to an $N$ -turn coil by multiplying by $N$ .
Explain, using the cancellation of perpendicular components between diametrically opposite current elements, why the axial field points only along the axis.
Use the right-hand rule to find the direction of the field along the axis from the current's circulation direction.
Recognise how the far-field limit $B \propto 1/x^3$ connects the loop to its identity as a magnetic dipole with moment $m = I\pi R^2$ .

Ask the tutor: Want to see the full off-axis geometry sketched out, or verify step by step that the axial formula smoothly turns into both the centre-field result and the dipole far-field limit?

Why does the field along the axis only ever point along the axis and never sideways?

It comes down to symmetry. Pick any tiny piece of the loop's current and look at the small magnetic field it produces at a point on the axis; then look at the piece exactly diametrically opposite it. Both pieces are the same distance from the axial point, so their contributions have the same size, but their sideways (perpendicular-to-axis) parts point in opposite directions from opposite sides of the loop, so they cancel exactly. Their along-the-axis parts, however, point the same way and add up. Do this for every pair of opposite elements around the loop, and every trace of a sideways field disappears, leaving a field that is purely along the axis.

Is the centre-field formula B equals mu-zero I over 2R just an approximation, or is it exact?

It is exact, not an approximation. It comes directly from setting $x = 0$ in the general formula $B = \mu_0 I R^2/[2(R^2+x^2)^{3/2}]$ , which itself is an exact result of integrating the Biot-Savart law around the loop with no simplifying assumptions. The far-away form, $B \approx \mu_0 I R^2/(2x^3)$ , is the one that is approximate, since it drops the $R^2$ term compared with $x^2$ and is only valid when $x$ is much larger than $R$ .

How is this axial-field formula actually used later in the chapter?

Two ways. First, taking the far-away limit shows the loop's field matches the field of a magnetic dipole with moment $m = I\pi R^2$ , which is exactly why a current loop is treated as a magnetic dipole in the next topic, with its own torque and potential-energy formulas in an external field. Second, the centre-field expression, generalised to $N$ turns, reappears directly inside the working of a moving-coil galvanometer, where the coil's own field interacts with a radial magnetic field from a permanent magnet to produce a deflecting torque.

My textbook sometimes writes x squared plus R squared and sometimes R squared plus x squared under the three-halves power — does the order matter?

No, addition is commutative, so $R^2+x^2$ and $x^2+R^2$ are exactly the same number and the formula is identical either way. What matters is not swapping the addition for the wrong combination altogether — writing $(R+x)^2$ instead of $R^2+x^2$ is a genuine error, not a harmless reordering, because $(R+x)^2 = R^2+2Rx+x^2$ includes an extra cross term that has no place in this formula. Always add the squares separately before taking the power.