Magnetic Field on the Axis of a Circular Current Loop

Moving Charges and Magnetism · CBSE

Theory

The set-up

A circular loop of radius $R$ carries a steady current $I$ . Place it in the $y$ – $z$ plane with its centre at the origin $O$ , so the $x$ -axis is the axis of the loop. The field is wanted at the point $P$ on the axis, at distance $x$ from $O$ .
Take a conducting element $d\vec{l}$ on the loop; let $\vec{s}$ be the vector from the element to $P$ , with $s^{2} = R^{2} + x^{2}$ .

The derivation CBSE sets (5 marks)

By Biot-Savart's law, $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\,\sin 90^\circ}{s^{2}} = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{\left(R^{2}+x^{2}\right)}$ The angle is $90^\circ$ because every element of the loop is perpendicular to the vector from it to an axial point: $d\vec{l}$ lies in the $y$ – $z$ plane while $\vec{s}$ lies in a plane containing the $x$ -axis, so $\lvert d\vec{l}\times\vec{s}\rvert = s\,dl$ .
$d\vec{B}$ is perpendicular to the plane of $d\vec{l}$ and $\vec{s}$ . Resolve it into a component along the axis and a component perpendicular to the axis.
The perpendicular components cancel in pairs: the contribution of any element is cancelled by that of the diametrically opposite element. Only the axial components survive, and they all add.
Writing $\phi$ for the angle whose sine projects $d\vec{B}$ onto the axis, $\sin\phi = \dfrac{R}{\left(R^{2}+x^{2}\right)^{1/2}}$ (equivalently, $\cos\theta = R/(x^2+R^2)^{1/2}$ in NCERT's labelling of the same projection).
Hence $B = \oint dB\sin\phi = \displaystyle\int_{0}^{2\pi R}\dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{\left(R^{2}+x^{2}\right)}\cdot\dfrac{R}{\left(R^{2}+x^{2}\right)^{1/2}}$
$R$ and $x$ are constants round the loop, so the integral is just $\displaystyle\int dl = 2\pi R$ : $B = \dfrac{\mu_0 I R}{4\pi\left(R^{2}+x^{2}\right)^{3/2}}\times 2\pi R$

$\boxed{\;B = \dfrac{\mu_0 I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}}\;}$

In vector form $\vec{B} = \dfrac{\mu_0 I R^{2}}{2\left(x^{2}+R^{2}\right)^{3/2}}\hat{i}$ , directed along the axis, its sense given by the right-hand thumb rule.
For a tightly wound coil of $N$ turns, multiply by $N$ : $B = \dfrac{\mu_0 N I R^{2}}{2\left(R^{2}+x^{2}\right)^{3/2}}$ .

Special cases

At the centre, $x = 0$ : $B = \dfrac{\mu_0 I}{2R} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I}{R}, \qquad B_{N\ \text{turns}} = \dfrac{\mu_0 N I}{2R}$
Far away on the axis, $x\gg R$ : drop $R^{2}$ from the denominator, $B \simeq \dfrac{\mu_0 I R^{2}}{2x^{3}} = \dfrac{\mu_0 I A}{2\pi x^{3}} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2m}{x^{3}}, \qquad m = IA = I\pi R^{2}$ the field of a magnetic dipole.
The ratio the board likes: with $B_1$ at the centre and $B_2$ on the axis at $x = R$ , $\dfrac{B_1}{B_2} = \dfrac{\mu_0 I/2R}{\mu_0 I R^{2}/2(2R^{2})^{3/2}} = 2\sqrt{2}$

Diagram in words

Draw the loop edge-on as an ellipse with its centre $O$ and the axis running horizontally to $P$ . Mark $R$ from $O$ to a current element $d\vec{l}$ at the top of the loop, and $\vec{s}$ from that element to $P$ .
At $P$ draw $d\vec{B}$ perpendicular to $\vec{s}$ , and resolve it into $dB\sin\phi$ along the axis and $dB\cos\phi$ perpendicular to it. Draw the diametrically opposite element's contribution too, so the two perpendicular components visibly cancel.
Field lines of the loop: closed curves passing through the centre along the axis, straight and closely spaced there, bending round outside the loop and closing on themselves — the same pattern as a short bar magnet, with the face from which the lines emerge acting as the north pole.

Right-hand thumb rule for a loop

Curl the palm of the right hand round the wire with the fingers pointing along the current; the outstretched thumb gives the direction of $\vec{B}$ on the axis.
Note this is a *different* rule from the one for a straight wire, where the thumb goes along the current and the fingers curl with the field. Fingers and thumb swap roles.

Combining loops — the recurring board problem

Two identical coils lying in perpendicular planes with a common centre produce fields at that centre which are mutually perpendicular, so they combine as $B_{\text{net}} = \sqrt{B_1^{2}+B_2^{2}}, \qquad \tan\alpha = \dfrac{B_2}{B_1}$
Radius $R$ , currents $I$ and $\sqrt{3}I$ : $B = \dfrac{\mu_0 I}{2R}\sqrt{1+3} = \dfrac{\mu_0 I}{R}$ , at $\tan^{-1}\left(1/\sqrt{3}\right) = 30^\circ$ to the field of the stronger coil.
Radius $5\ \text{cm}$ , currents $3\ \text{A}$ and $4\ \text{A}$ : $B_{\text{net}} = \dfrac{\mu_0}{10^{-1}}\sqrt{9+16} = \dfrac{5\mu_0}{10^{-1}} = 62.8\ \mu\text{T}$ , making $\tan^{-1}(3/4) = 37^\circ$ with the field of the $3\ \text{A}$ coil.
Two small identical loops with perpendicular axes meeting at $O$ , each centre a distance $x$ from $O$ : each contributes $B = \dfrac{\mu_0}{4\pi}\dfrac{2\pi I a^{2}}{\left(a^{2}+x^{2}\right)^{3/2}}$ and, being perpendicular, they give $B_{\text{net}} = \sqrt{2}\,B_1$ at $45^\circ$ to each.

Composite arrangements at a point

Break the conductor into parts, compute each contribution with the right standard result, fix each direction with the right-hand rule, then add as signed quantities (or as vectors).
Straight wire bent into a loop in the middle: at the centre $O$ , the straight parts give $\dfrac{\mu_0}{4\pi}\dfrac{2I}{R}$ and the loop gives $\dfrac{\mu_0}{4\pi}\dfrac{2\pi I}{R}$ , opposite in sense, so $B_{\text{net}} = \dfrac{\mu_0 I}{2\pi R}\left(\pi - 1\right) = \dfrac{\mu_0 I}{2R}\left(1-\dfrac{1}{\pi}\right)$
Semicircular arc of $2\ \text{A}$ and radius $12\ \text{cm}$ together with a small loop of $1\ \text{A}$ and radius $4\ \text{cm}$ : $B_{\text{arc}} = \dfrac{\mu_0 I}{4r} = 0.523\times10^{-5}\ \text{T}$ outwards, $B_{\text{loop}} = \dfrac{\mu_0 I}{2r'} = 1.57\times10^{-5}\ \text{T}$ inwards, so $B_{\text{net}} = 1.047\times10^{-5}\ \text{T}$ directed into the plane of the paper.

Results to carry away

Axial field $\dfrac{\mu_0 I R^{2}}{2(R^{2}+x^{2})^{3/2}}$ ; centre $\dfrac{\mu_0 N I}{2R}$ ; far field $\dfrac{\mu_0}{4\pi}\dfrac{2m}{x^{3}}$ .
$B_{\text{centre}}/B_{\text{axis at }x=R} = 2\sqrt{2}$ .
Perpendicular coils at a common centre: add by Pythagoras.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Two identical circular loops $P$ and $Q$ , each of radius $R$ carrying current $I$ are kept in perpendicular planes such that they have a common centre $O$ (loop $Q$ oriented vertically, loop $P$ horizontally). Find the magnitude and direction of the net magnetic field at point $O$ .
2 marksCBSE Class 12 Physics — SA · 2023 · Section B, 2 marksAnswer 1 ↓
2
Two very small identical circular loops, (1) and (2), carrying equal currents $I$ are placed vertically (with respect to the plane of the paper) with their geometrical axes perpendicular to each other, meeting at a common point $O$ , each loop's centre being at a distance $x$ from $O$ along its own axis. Find the magnitude and direction of the net magnetic field produced at the point $O$ .
2 marksCBSE Class 12 Physics — SA · 2017C, Foreign 2014, Ap · Section B, 2 marksAnswer 2 ↓
3
A current carrying circular loop and a straight wire bent partly in the form of a semicircle are placed as shown in the figure. Find the magnitude and direction of net magnetic field at point $O$ .
3 marksCBSE Class 12 Physics — SA · 2023, Ev · Section C, 3 marksAnswer 3 ↓
4
(a) State Biot-Savart's law and express this law in the vector form. (b) Two identical circular coils, $P$ and $Q$ each of radius $R$ , carrying currents 1 A and $\sqrt{3}$ A respectively, are placed concentrically and perpendicular to each other lying in the XY and YZ planes. Find the magnitude and direction of the net magnetic field at the centre of the coils.
5 marksCBSE Class 12 Physics — Long Answer · AI 2017, Ev · Section E, 5 marksAnswer 4 ↓
5
Two identical loops $P$ and $Q$ each of radius 5 cm are lying in perpendicular planes (loop $Q$ vertical, loop $P$ horizontal) such that they have a common centre $O$ . Find the magnitude and direction of the net magnetic field at the common centre of the two coils, if they carry currents equal to 3 A (loop $P$ ) and 4 A (loop $Q$ ) respectively.
5 marksCBSE Class 12 Physics — Long Answer · AI 2017, Ap · Section E, 5 marksAnswer 5 ↓
6
Use Biot-Savart law to derive the expression for the magnetic field on the axis of a current carrying circular loop of radius $R$ . Draw the magnetic field lines due to circular wire carrying current $I$ .
5 marksCBSE Class 12 Physics — Long Answer · AI 2016 · Section E, 5 marksAnswer 6 ↓
7
Two identical coils $P$ and $Q$ each of radius $R$ are lying in perpendicular planes such that they have a common centre. Find the magnitude and direction of the magnetic field at the common centre of the two coils, if they carry currents equal to $I$ and $\sqrt{3}\,I$ respectively.
5 marksCBSE Class 12 Physics — Long Answer · Foreign 2016 · Section E, 5 marksAnswer 7 ↓
8
A circular loop of radius $R$ carries a current $I$ . Obtain an expression for the magnetic field at a point on its axis at a distance $x$ from its centre. OR Write, using Biot-Savart law, the expression for the magnetic field $\vec{B}$ due to an element $d\vec{\ell}$ carrying current $I$ at a distance $\vec{r}$ from it in a vector form. Hence derive the expression for the magnetic field due to a current carrying loop of radius $R$ at a point $P$ distant $x$ from its centre along the axis of the loop.
5 marksCBSE Class 12 Physics — Long Answer · 3/5, 2020 · Section E, 5 marksAnswer 8 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

9
The magnetic field at the centre of a current carrying circular loop of radius $R$ is $B_1$ . The magnetic field at a point on its axis at a distance $R$ from the centre of the loop is $B_2$ . Then the ratio $(B_1/B_2)$ is
- (a)
  $2\sqrt{2}$
- (b)
  $\dfrac{1}{\sqrt{2}}$
- (c)
  $\sqrt{2}$
- (d)
  $2$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 9 ↓
10
A current $I$ flows through a long straight conductor which is bent into a circular loop of radius $R$ in the middle (the straight portions continue on either side, and the centre of the loop is point $O$ ). The magnitude of the net magnetic field at point $O$ will be
- (a)
  Zero
- (b)
  $\dfrac{\mu_0 I}{2R}(1+\pi)$
- (c)
  $\dfrac{\mu_0 I}{4\pi R}$
- (d)
  $\dfrac{\mu_0 I}{2R}\left(1-\dfrac{1}{\pi}\right)$
1 markCBSE Class 12 Physics — Multiple Choice · AI 2020 · Section A, 1 markAnswer 10 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
Let us assume that the current in loop $Q$ flows such that there is equal current flow in each parallel wire of loop $Q$ (i.e. the two halves of loop $Q$ , as seen from the point of interest which lies along its own diameter/axis in the plane, produce cancelling fields there). Since the point of interest (the common centre $O$ itself, lying in the plane of loop $Q$ along a diameter direction relevant here) is such that the magnetic field due to loop $Q$ 's own plane geometry at $O$ effectively vanishes by the stated symmetry, the magnetic field due to $Q$ at $O$ is zero: $B_Q = 0$ . Now, current in loop $P$ is $I$ , so magnetic field due to current $I$ at the centre of loop $P$ : $B_P = \dfrac{\mu_0 I}{2R}$ Hence, total magnetic field at $O = B_P + B_Q = \dfrac{\mu_0 I}{2R}$ , directed perpendicular into the plane of loop $P$ .
2.↑ back to the question
The magnetic field at an axial point due to a circular loop is given by $B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I a^2}{(a^2+r^2)^{3/2}}$ , where $I$ is the current, $a$ is the radius of the loop and $r$ is the distance of $O$ from the centre of the loop. Since $I$ , $a$ and $r = x$ are the same for both loops, the magnitude of $B$ will be the same for each and is given by $B_1 = B_2 = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I a^2}{(a^2+x^2)^{3/2}}$ The direction of the magnetic field due to loop (1) will be away from $O$ along its axis, and that due to loop (2) will be towards $O$ along its own (perpendicular) axis. Since $B_1$ and $B_2$ are perpendicular to each other (loop axes are at 90°), The magnitude of the net magnetic field is given by $B_{net} = \sqrt{B_1^2 + B_2^2} = \sqrt{2}\,B_1 = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\sqrt{2}\pi I a^2}{(a^2+x^2)^{3/2}}$ directed at 45° to each of $B_1$ and $B_2$ (i.e. bisecting the angle between the two loop axes).
3.↑ back to the question
The magnetic field induction at point $O$ due to the semicircular arc (current 2 A, radius $r = 12$ cm) is $B = \dfrac{\mu_0 I \pi}{4\pi r} = \dfrac{\mu_0 I}{4r} = \dfrac{4\pi\times10^{-7}\times2}{4\times12\times10^{-2}} = 0.523\times10^{-5}$ T The direction of $B$ , according to the right hand rule, is normal to the plane of paper, directed outwards. The magnetic field induction at the centre of the smaller circular loop (current 1 A, radius $r' = 4$ cm) is given by $B' = \dfrac{\mu_0}{4\pi}\times\dfrac{2\pi I}{r'} = \dfrac{\mu_0 I}{2r'} = \dfrac{4\pi\times10^{-7}\times1}{2\times4\times10^{-2}} = 1.57\times10^{-5}$ T The direction of $B'$ , according to the right hand rule, is perpendicular to the plane of paper and directed inwards. Now, net magnetic field, $B_{net} = B' - B = 1.57\times10^{-5} - 0.523\times10^{-5} = 1.047\times10^{-5}$ T (inwards)
4.↑ back to the question
(a) A current carrying wire produces a magnetic field around it. Biot-Savart's law states that the magnitude of intensity of small magnetic field $d\vec{B}$ due to current $I$ carrying element $d\vec{\ell}$ at any point $P$ at distance $r$ from it is given by $|d\vec{B}| = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell\sin\theta}{r^2}$ , where $\theta$ is the angle between $\vec{r}$ and $d\vec{\ell}$ . In vectorial form: $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{\ell}\times\vec{r}}{r^3}$ (b) Field due to current in coil $P$ (in the XY plane, current 1 A) is $\vec{B}_1 = \dfrac{\mu_0 I_1}{2R}\hat{k}$ ; field due to coil $Q$ (in the YZ plane, current $\sqrt{3}$ A) is $\vec{B}_2 = \dfrac{\mu_0 I_2}{2R}\hat{i}$ $\therefore \vec{B} = \vec{B}_1 + \vec{B}_2 = \left(\dfrac{\mu_0}{2R}\right)\hat{k} + \left(\dfrac{\sqrt{3}\mu_0}{2R}\right)\hat{i}$ (since $I_1 = 1$ A, $I_2 = \sqrt{3}$ A) $|\vec{B}| = \sqrt{\left(\dfrac{\mu_0}{2R}\right)^2 + \left(\dfrac{\sqrt{3}\mu_0}{2R}\right)^2} = \dfrac{\mu_0}{2R}\sqrt{1+3} = \dfrac{\mu_0}{2R}\times2$ $|\vec{B}| = \dfrac{\mu_0}{R}$ , and the resultant magnetic field is directed in the XZ plane.
5.↑ back to the question
Magnetic field induction due to vertical loop $Q$ (current $I_2 = 4$ A) at the centre $O$ is: $B_2 = \dfrac{\mu_0 I_2}{2R} = \dfrac{4\mu_0}{10^{-1}}$ (with $R = 5$ cm $= 5\times10^{-2}$ m) Magnetic field induction due to horizontal loop $P$ (current $I_1 = 3$ A) at the centre $O$ is: $B_1 = \dfrac{\mu_0 I_1}{2R} = \dfrac{3\mu_0}{10^{-1}}$ Since $B_1$ and $B_2$ are perpendicular to each other, the resultant magnetic field induction at the centre $O$ is: $B_{net} = \sqrt{B_1^2+B_2^2} = \sqrt{\left(\dfrac{4\mu_0}{10^{-1}}\right)^2 + \left(\dfrac{3\mu_0}{10^{-1}}\right)^2} = \dfrac{\mu_0}{10^{-1}}\sqrt{9+16} = \dfrac{5\mu_0}{10^{-1}}$ $= 50\times4\pi\times10^{-7} = 62.8\times10^{-6}$ T $= 62.8\ \mu$ T Direction of resultant magnetic field: $\tan\theta = \dfrac{B_2}{B_1} = \dfrac{3\mu_0\times10^{-1}}{4\mu_0\times10^{-1}}$ [using ratio $I_1{:}I_2 = 3{:}4$ ] $\tan\theta = \dfrac{3}{4}$ or $\theta = 37°$ So the resultant magnetic field $B$ makes an angle 37° with $B_1$ (the field of the 3 A loop).
6.↑ back to the question
Consider a circular loop of radius $R$ carrying current $I$ , and a point $P$ on its axis at distance $x$ from the centre $O$ . By Biot-Savart's law, small field due to element $d\vec{\ell}$ at $P$ : $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell\sin90°}{s^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}$ , since $d\vec{\ell} \perp \vec{s}$ everywhere on the loop (where $s = \sqrt{R^2+x^2}$ ). The component $dB\cos\phi$ (perpendicular to the axis) due to a current element is cancelled by the equal and opposite component $dB\cos\phi$ of the diametrically opposite current element, whereas the components $dB\sin\phi$ (along the axis) add up. So, net field at $P$ due to entire loop is $B = \oint dB\sin\phi = \displaystyle\int_0^{2\pi R}\dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}\cdot\dfrac{R}{(R^2+x^2)^{1/2}}$ $B = \dfrac{\mu_0 IR}{4\pi(R^2+x^2)^{3/2}}\int_0^{2\pi R}d\ell = \dfrac{\mu_0 IR}{4\pi(R^2+x^2)^{3/2}}\times2\pi R$ $B = \dfrac{\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$ , directed along the axis: (a) towards the coil if current in it is anticlockwise, or (b) away from the coil if current is clockwise (as seen from that side) — [book states: towards the coil if current is clockwise, away if anticlockwise, per the sign convention used]. The magnetic field lines due to a circular wire carrying current $I$ form closed loops that pass through the centre along the axis (straight and dense near the centre) and curve around and close up outside the loop, resembling those of a bar magnet's field, being directed by the right-hand thumb rule.
7.↑ back to the question
Since both coils are inclined to each other at an angle of 90°, the magnitude of their resultant magnetic field at the common centre will be $B = \sqrt{B_P^2+B_Q^2} = \dfrac{\mu_0 I}{2R}\sqrt{1+3} = \dfrac{\mu_0 I}{R}$ (using $B_P = \dfrac{\mu_0 I}{2R}$ and $B_Q = \dfrac{\mu_0\sqrt{3}I}{2R}$ , perpendicular to each other) The direction of the resultant field is at an angle $\theta$ given by $\theta = \tan^{-1}\left(\dfrac{B_P}{B_Q}\right) = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right) = 30°$ Hence, the direction of the magnetic field will be at an angle 30° to the plane of loop $Q$ (i.e. 30° from $B_Q$ 's direction, towards $B_P$ 's direction).
8.↑ back to the question
By Biot-Savart's law, the small magnetic field due to a current element of the circular coil of radius $R$ at point $P$ (on the axis, distance $x$ from centre) is $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell\sin90°}{s^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}$ , since $d\vec{\ell} \perp \vec{s}$ , where $s = \sqrt{R^2+x^2}$ . The component $dB\cos\phi$ (perpendicular to axis) cancels for diametrically opposite elements; the axial components $dB\sin\phi$ add up. $B = \oint dB\sin\phi = \displaystyle\int_0^{2\pi R}\dfrac{\mu_0}{4\pi}\dfrac{I\,d\ell}{(R^2+x^2)}\cdot\dfrac{R}{(R^2+x^2)^{1/2}} = \dfrac{\mu_0 IR}{4\pi(R^2+x^2)^{3/2}}\times2\pi R$ $B = \dfrac{\mu_0 IR^2}{2(R^2+x^2)^{3/2}}$ , directed along the axis (towards the coil if current is clockwise, away if anticlockwise, as seen from the point of observation).
9.Answer: (A)↑ back to the question
Answer: (a) $2\sqrt{2}$ Magnetic field due to loop at its centre: $B_1 = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I}{R}$ ...(i) Magnetic field due to a current carrying loop, on its axis at distance $x$ : $B = \dfrac{\mu_0}{2}\cdot\dfrac{Ix^2}{(R^2+x^2)^{3/2}}$ Putting $x = R$ : $B_2 = \dfrac{\mu_0}{2}\cdot\dfrac{IR^2}{(2R^2)^{3/2}} = \dfrac{\mu_0}{2}\cdot\dfrac{IR^2}{2^{3/2}R^3} = \dfrac{\mu_0 I}{4\sqrt{2}R}$ ...(ii) Dividing (i) by (ii): $\dfrac{B_1}{B_2} = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I \times 4\sqrt{2}R}{R \cdot I \mu_0} = 2\sqrt{2}$
10.Answer: (D)↑ back to the question
Answer: (d) $\dfrac{\mu_0 I}{2R}\left(1-\dfrac{1}{\pi}\right)$ Magnetic field due to straight wire at $O$ : $B_1 = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{R}$ (out of the page, say) Magnetic field due to circular wire (loop) at $O$ : $B_2 = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2\pi I}{R}$ (into the page) $B_{net} = B_2 - B_1 = \dfrac{\mu_0}{4\pi R}[2\pi I - 2I] = \dfrac{\mu_0 I}{4\pi R}\times2(\pi - 1) = \dfrac{\mu_0 I}{2\pi R}(\pi-1)$ $B_{net} = \dfrac{\mu_0 I}{2R}\left(1 - \dfrac{1}{\pi}\right)$