The value of magnetic field at point O in the given figure is: (A) $\dfrac{\mu_0 I}{2\pi R}$ (B) $\dfrac{\mu_0 I}{\pi R}$ (C) $\dfrac{\mu_0 I}{4R}$ (D) $\dfrac{\mu_0 I}{R}$ ![Figure — 55/6/1 Q3](/figures/pyq/2025-55-6-1/fig-q3.svg)

The value of magnetic field at point O in the given figure is: (A) (μ₀ I)/(2π R) (B) (μ₀ I)/(π R) (C) (μ₀ I)/(

Question

\dfrac{\mu_0 I}{2\pi R}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Magnetic Field of a Circular Loop

Magnetic Field of a Circular Loop — From Intuition to Formula

Imagine you take a piece of wire and bend it into a perfect circle. Now push a steady current through it. What happens to the space around it? The moving charges in every tiny segment of the wire create their own little magnetic fields. The question is: what do all these tiny fields add up to?

The Intuition: Symmetry and Superposition

At the centre of the loop, something beautiful happens. Every small piece of the wire is at the same distance $R$ from the centre. And each piece sends its magnetic field contribution straight toward (or away from) the centre, depending on the current direction. Because the loop is symmetric, the sideways components from opposite sides cancel out perfectly. Only the component perpendicular to the plane of the loop survives.

So the net field at the centre points straight along the axis of the loop — either out of the page or into it, following the right-hand rule.

Tip

Right-hand rule for the loop

Curl the fingers of your right hand in the direction of the current. Your thumb points in the direction of the magnetic field at the centre.

The Precise Statement

For a circular loop of radius $R$ carrying a steady current $I$ , the magnetic field at the centre is:

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

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

The field is uniform only at the exact centre — move even a little along the axis, and the magnitude changes. But at the centre, the formula is exact.

B_{\text{centre}} = \frac{\mu_0 I}{2R}

Where Does This Come From?

You can derive it using the Biot–Savart law. Each current element $I\,d\ell$ produces a field:

dB = \frac{\mu_0}{4\pi} \frac{I\,d\ell \sin\theta}{r^2}

At the centre, every element is perpendicular to the radius vector ( $\theta = 90^\circ$ , so $\sin\theta = 1$ ), and the distance $r = R$ is constant. The direction of each $dB$ is the same (along the axis). So you just integrate $d\ell$ around the full circumference $2\pi R$ :

B = \int dB = \frac{\mu_0 I}{4\pi R^2} \int_0^{2\pi R} d\ell = \frac{\mu_0 I}{4\pi R^2} \cdot 2\pi R = \frac{\mu_0 I}{2R}

Note

The $4\pi$ in the denominator of the Biot–Savart law cancels with the $2\pi$ from the circumference, leaving the clean $2R$ in the denominator.

What If You Have Multiple Turns?

If you wind the wire into $N$ closely spaced turns (a flat coil), each turn contributes the same field at the centre. The fields simply add:

B = \frac{\mu_0 N I}{2R}

This is why a coil of many turns produces a stronger field — each turn is an independent source.

Watch out

Common mistake

Do not confuse this with the field on the axis at a distance $x$ from the centre. That formula is $B = \frac{\mu_0 I R^2}{2(R^2 + x^2)^{3/2}}$ , which reduces to $\frac{\mu_0 I}{2R}$ only when $x = 0$ . Many students mistakenly use the axis formula for the centre — it works, but it's overkill. The centre formula is simpler and exact.

Why This Matters

The circular loop is the building block of solenoids and electromagnets. Understanding the field at its centre gives you the foundation for designing coils that produce uniform magnetic fields — used in everything from MRI machines to particle accelerators.

Final answer: At the centre of a circular loop of radius $R$ carrying current $I$ , the magnetic field is $B = \frac{\mu_0 I}{2R}$ , directed along the axis of the loop.

The magnetic field at the centre of a circular current loop is a classic CBSE Class 12 Physics NCERT formula, often searched as magnetic field at the centre of a circular loop derivation or moving charges and magnetism important questions class 12. Because this same setup extends into axial-field and solenoid problems tested in JEE Main and NEET, it is worth deriving from the Biot-Savart law at least once rather than only memorising the final expression.

The magnetic field at the centre of a semicircular current-carrying arc is found using the Biot-Savart law; each element contributes perpendicular to the plane, and integration over the arc gives $\boxed{\frac{\mu_0 I}{4R}}$ .

The magnetic field at a point due to a current-carrying conductor is governed by the Biot-Savart law. For symmetric geometries like the circular arc shown in the figure, the calculation simplifies beautifully because every current element contributes in the same direction at the centre.

When a current $I$ flows through a circular arc, each infinitesimal element $d\vec{l}$ produces a magnetic field at the centre that points perpendicular to the plane of the arc (determined by the right-hand rule). The key insight is that for a circular arc of radius $R$ , the perpendicular distance from every element to the centre is exactly $R$ , and the angle between $d\vec{l}$ and the position vector is always $90°$ .

\theta

Let me show you why this formula works and apply it to our semicircle.

Apply the Biot-Savart law to a circular arc element For an element $d\vec{l}$ on the arc at distance $R$ from point O:

dB = \frac{\mu_0 I}{4\pi} \frac{|d\vec{l} \times \vec{r}|}{r^3} = \frac{\mu_0 I}{4\pi R^2} dl

Since $dl = R \, d\theta$ (arc length element), we have:

dB = \frac{\mu_0 I}{4\pi R} d\theta

Integrate over the semicircular arc A semicircle subtends an angle $\theta = \pi$ radians at its centre. All field contributions point in the same direction (into or out of the page), so:

B = \int_0^{\pi} \frac{\mu_0 I}{4\pi R} d\theta = \frac{\mu_0 I}{4\pi R} \cdot \pi = \frac{\mu_0 I}{4R}

Check the straight segments The straight wire segments connecting to the semicircle pass through (or point toward/away from) point O. For any straight wire element along the line joining it to the field point, $d\vec{l}$ is parallel to $\vec{r}$ , making $d\vec{l} \times \vec{r} = 0$ . These segments contribute zero field at O.

Tip

For a full circular loop, $\theta = 2\pi$ , giving $B = \frac{\mu_0 I}{2R}$ at the centre. A semicircle is exactly half of this.

Watch out

Don't confuse this with the field of a straight wire at perpendicular distance $R$ , which is $\frac{\mu_0 I}{2\pi R}$ . The geometry matters crucially.

✓Final answer

\dfrac{\mu_0 I}{4R}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 14 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. An arc of a circle of radius R subtends an angle π/2 at the centre. It carries a current i. The magnetic induction at the centre will be$
(a) μ0 i / 2R
(b) μ0 i / 8R
(c) μ0 i / 4R
(d) 2μ0 i / 5R
›Reveal solutionSolution

The magnetic induction at the centre due to a current-carrying arc subtending angle $\theta$ is $\dfrac{\mu_0 i}{4\pi R}\theta$ ; for $\theta=\pi/2$ this gives $\mu_0 i/(8R)$ .

For a full circular loop of radius $R$ carrying current $i$ , the magnetic field at the centre is $B_{\text{full}} = \dfrac{\mu_0 i}{2R}$ , using the Biot–Savart law integrated over the full $2\pi$ angle.

For an arc subtending angle $\theta$ at the centre, the field is proportional to $\theta$ (a fraction $\theta/2\pi$ of the full loop's contribution):

$B = \dfrac{\mu_0 i}{2R}\times\dfrac{\theta}{2\pi} = \dfrac{\mu_0 i\,\theta}{4\pi R}$

With $\theta = \pi/2$ :

$B = \dfrac{\mu_0 i}{4\pi R}\times\dfrac{\pi}{2} = \dfrac{\mu_0 i}{8R}$

✓Final answer
$\mu_0 i/(8R)$
CBSE 2025Set 55/6/11 markMCQ
$\dfrac{\mu_0 I}{2\pi R}$
›Reveal solutionSolution
The magnetic field at the centre of a semicircular current-carrying arc is found using the Biot-Savart law; each element contributes perpendicular to the plane, and integration over the arc gives $\boxed{\frac{\mu_0 I}{4R}}$ .

Figure — 55/6/1 Q3

The magnetic field at a point due to a current-carrying conductor is governed by the Biot-Savart law. For symmetric geometries like the circular arc shown in the figure, the calculation simplifies beautifully because every current element contributes in the same direction at the centre.

When a current $I$ flows through a circular arc, each infinitesimal element $d\vec{l}$ produces a magnetic field at the centre that points perpendicular to the plane of the arc (determined by the right-hand rule). The key insight is that for a circular arc of radius $R$ , the perpendicular distance from every element to the centre is exactly $R$ , and the angle between $d\vec{l}$ and the position vector is always $90°$ .
$\theta$
Let me show you why this formula works and apply it to our semicircle.

Apply the Biot-Savart law to a circular arc element For an element $d\vec{l}$ on the arc at distance $R$ from point O:

$dB = \frac{\mu_0 I}{4\pi} \frac{|d\vec{l} \times \vec{r}|}{r^3} = \frac{\mu_0 I}{4\pi R^2} dl$
Since $dl = R \, d\theta$ (arc length element), we have:
$dB = \frac{\mu_0 I}{4\pi R} d\theta$

Integrate over the semicircular arc A semicircle subtends an angle $\theta = \pi$ radians at its centre. All field contributions point in the same direction (into or out of the page), so:

$B = \int_0^{\pi} \frac{\mu_0 I}{4\pi R} d\theta = \frac{\mu_0 I}{4\pi R} \cdot \pi = \frac{\mu_0 I}{4R}$

Check the straight segments The straight wire segments connecting to the semicircle pass through (or point toward/away from) point O. For any straight wire element along the line joining it to the field point, $d\vec{l}$ is parallel to $\vec{r}$ , making $d\vec{l} \times \vec{r} = 0$ . These segments contribute zero field at O.

Tip

For a full circular loop, $\theta = 2\pi$ , giving $B = \frac{\mu_0 I}{2R}$ at the centre. A semicircle is exactly half of this.

Watch out

Don't confuse this with the field of a straight wire at perpendicular distance $R$ , which is $\frac{\mu_0 I}{2\pi R}$ . The geometry matters crucially.

✓Final answer
$\dfrac{\mu_0 I}{4R}$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The magnetic field at the centre of a circular coil of radius 5 cm carrying a current of 1 A is 1.256 T. If the radius is made 10 cm, then the magnetic field at the centre of the loop carrying the same current will be$
(a) 2.512 T
(b) 0.628 T
(c) 5.024 T
(d) 0.314 T
›Reveal solutionSolution

For a circular coil, $B_{centre} = \mu_0 I/(2R)$ , so $B \propto 1/R$ at fixed current; doubling $R$ halves $B$ , giving $0.628$ T.

Formula for field at the centre of a circular coil

$B = \frac{\mu_0 I}{2R}$

For a fixed current $I$ , $B$ is inversely proportional to the radius $R$ :

$\frac{B_2}{B_1} = \frac{R_1}{R_2}$

Substituting the given values

$R_1 = 5$ cm, $B_1 = 1.256$ T, $R_2 = 10$ cm (current unchanged at 1 A):

$B_2 = B_1 \times \frac{R_1}{R_2} = 1.256 \times \frac{5}{10} = 0.628\ \text{T}$

Note on the printed value: Using $\mu_0 = 4\pi\times10^{-7}$ T $\cdot$ m/A, $I=1$ A and $R=0.05$ m, the actual field at the centre works out to $B = \mu_0 I/(2R) \approx 1.256\times10^{-5}$ T, not $1.256$ T -- the value printed in the question is larger by a factor of $10^5$ . This does not affect the answer to this question, since only the ratio $B\propto 1/R$ is required, and that ratio law is applied to the number as given.

✓Final answer
$0.628$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Due to 10 amperes of current flowing in a circular coil of 10 cm radius, the magnetic field produced at its centre is 3.14 x 10^-3 weber/m^2. The number of turns in the coil will be$
(i) 5000
(ii) 100
(iii) 50
(iv) 25
›Reveal solutionSolution

N = B(2r)/(mu0 I) = 50 turns.

The field at the centre of a coil of N turns is $B = \frac{\mu_0 N I}{2r}$ . Solving for N:

$N = \frac{B \cdot 2r}{\mu_0 I} = \frac{(3.14\times10^{-3})(2\times0.10)}{(4\pi\times10^{-7})(10)} = \frac{6.28\times10^{-4}}{1.257\times10^{-5}} \approx 50.$

✓Final answer
$(iii) 50.$
CBSE 2024Set A1 markMCQ
$Q. The magnetic field produced at the centre of current carrying circular coil is (A) on the plane of coil (B) perpendicular to the plane of coil (C) at 45° to the plane of coil (D) at 180° to the plane of coil$
›Reveal solutionSolution

At the centre the magnetic field points along the axis — perpendicular to the plane of the coil.

Using the Biot–Savart law, every current element of a circular coil contributes a field at the centre that points along the axis of the coil (perpendicular to its plane); by symmetry the in-plane components cancel. The magnitude at the centre is

$B = \frac{\mu_0 N I}{2R}.$

The direction is given by the right-hand rule and is perpendicular to the plane of the coil, along its axis.

✓Final answer
$(B) perpendicular to the plane of coil.$
CBSE 2024Set ANNUAL1 mark
$12.0\ \Omega$
›Reveal solutionSolution

A semicircular arc contributes exactly half the field of a full circular loop of the same radius at the centre, since the field from a circular arc is proportional to the angle it subtends. The alternative converts a galvanometer to an ammeter using a low-resistance shunt in parallel.

Field at the centre of a semicircular arc

For a full circular loop of radius $R$ carrying current $I$ , the field at the centre is $B_{loop} = \dfrac{\mu_0 I}{2R}$ . A semicircular arc subtends half the angle ( $\pi$ instead of $2\pi$ ) at the centre, so it contributes exactly half this field:

$B = \frac{\mu_0 I}{4R}$

Substituting $I = 10\,\text{A}$ , $R = 20\,\text{cm} = 0.20\,\text{m}$ :

$B = \frac{(4\pi\times10^{-7})(10)}{4(0.20)} = \frac{4\pi\times10^{-6}}{0.8} = \frac{12.566\times10^{-6}}{0.8} = 1.57\times10^{-5}\,\text{T}$

✓Final answer
$B \approx 1.57\times10^{-5}\,\text{T} = 15.7\,\mu\text{T}$

To convert a galvanometer into an ammeter of a much larger range, a low-resistance shunt $S$ is connected in parallel with the coil so that most of the current bypasses the coil, and only the small current $I_g$ needed for full-scale deflection flows through it.

Given: coil resistance $G = 12.0\,\Omega$ , full-scale deflection current $I_g = 2.5\,\text{mA} = 2.5\times10^{-3}\,\text{A}$ , desired range $I = 7.5\,\text{A}$ .

For the parallel combination, the potential difference across the galvanometer equals that across the shunt:

$I_g G = (I - I_g)S \implies S = \frac{I_g G}{I - I_g}$

$S = \frac{(2.5\times10^{-3})(12.0)}{7.5 - 2.5\times10^{-3}} = \frac{0.03}{7.4975} \approx 4.0\times10^{-3}\,\Omega$

✓Final answer
$4.0\times10^{-3}\,\Omega$
CBSE 2024Set ANNUAL1 markMCQ
$Q. An infinitely long straight conductor is bent into the shape as shown in the figure. The magnetic field at the centre of the circular part is -$
(a) (μ₀/4π)·(2πI/R)
(b) (μ₀/4π)·(2I/R)(π+1)
(c) (μ₀/4π)·(2I/R)(π−1)
(d) Zero
›Reveal solutionSolution

The field at the centre is the sum of the field from the full circular loop and the field from the two straight semi-infinite wire segments (which together act like one infinite straight wire tangent to the circle).

Field due to the circular loop (radius $R$ , full turn):

$B_{circle} = \dfrac{\mu_0 I}{2R}$

Field due to the straight parts: The two straight segments are collinear and tangent to the circle at the junction point, so together they behave like a single infinite straight wire at perpendicular distance $R$ from the centre:

$B_{straight} = \dfrac{\mu_0 I}{2\pi R}$

Both fields point in the same direction at the centre (same current sense), so they add:

$B = \dfrac{\mu_0 I}{2R} + \dfrac{\mu_0 I}{2\pi R} = \dfrac{\mu_0 I}{2R}\left(1+\dfrac{1}{\pi}\right) = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{R}(\pi+1)$

✓Final answer
$B = \dfrac{\mu_0}{4\pi}\cdot\dfrac{2I}{R}(\pi+1)$
CBSE 2023Set ANNUAL1 markMCQ
$Q. A circular coil of radius 10 cm having 100 turns carries a current of 3.2 A. The magnetic field at the centre of the coil is$
(1) 2.01 x 10^-3 T
(2) 5.64 x 10^-3 T
(3) 2.64 x 10^-4 T
(4) 5.64 x 10^-4 T
›Reveal solutionSolution

Use the formula for the magnetic field at the centre of a circular current loop, scaled by the number of turns.

$B=\dfrac{\mu_0 N I}{2R}=\dfrac{(4\pi\times10^{-7})(100)(3.2)}{2(0.1)}$

Numerator: $4\pi\times10^{-7}\times320=4.02\times10^{-4}$

$B=\dfrac{4.02\times10^{-4}}{0.2}=2.01\times10^{-3}\ \text{T}$

✓Final answer
$2.01\times10^{-3}$
CBSE 2022Set ANNUAL1 mark
$Q. Draw a diagram of the magnetic field lines due to a current carrying circular loop.$
›Reveal solutionSolution
Figure — Stem asks to draw the magnetic field lines due to a current-carrying circular loop; the catalog figure is prec

The magnetic field lines of a current-carrying circular loop form closed loops threading through the coil, straight along the central axis, and curving back around outside — the same pattern as the field of a bar magnet.

Description of the field-line pattern for a circular current loop (current flowing, say, anticlockwise when viewed from the front):

Near the wire itself, the field lines are small concentric circles around the wire (as for any straight current element), by the right-hand rule.

At the centre of the loop and along its axis, the field lines are straight, directed along the axis (perpendicular to the plane of the loop), all pointing the same way (say, out of the page if current is anticlockwise).

Away from the loop, the field lines curve around, emerging from one face of the loop, arcing outward, and re-entering the other face — exactly the same overall pattern as the magnetic field lines of a short bar magnet, so the loop behaves like a magnetic dipole with a north pole on the face from which the lines emerge and a south pole on the other face.

The lines are densest inside the loop (strongest field at the centre) and thin out with distance, forming closed loops (magnetic field lines never begin or end).

✓Final answer
$A pattern of closed, concentric loops threading the coil — straight along the axis, curving back outside — identical in form to a bar magnet's dipole field.$
CBSE 2022Set ANNUAL1 mark
$Q. Consider the circuit shown where APB and AQB are semicircles. What will be the magnetic field at the centre O of the circle?$
›Reveal solutionSolution

The two semicircular arcs carry equal currents in opposite senses about O, so their fields cancel: $B_O = 0$ .

The current $i$ entering at $A$ divides between the upper arc $APB$ and the lower arc $AQB$ . Since the two semicircles have the same radius and hence the same length and resistance, the current divides equally: each carries $i/2$ .

The magnetic field at the centre of a semicircular arc of radius $R$ carrying current $I$ is

$B = \frac{\mu_0 I}{4R}.$

So each arc gives a field of magnitude

$B_1 = B_2 = \frac{\mu_0 (i/2)}{4R} = \frac{\mu_0 i}{8R}.$

Crucially, the current in the upper arc circulates around $O$ in the opposite rotational sense to the current in the lower arc (one tends to produce a field into the page at $O$ , the other out of the page). Being equal in magnitude and opposite in direction, the two contributions cancel:

$B_O = B_1 - B_2 = 0.$

✓Final answer
$Zero — the fields of the two equal semicircular halves cancel exactly at the centre O.$
CBSE 2019Set ANNUAL1 markMCQ
$Q. Figure shows a circular coil with centre 'O'. If current flows through the coil, the magnetic field produced by it is maximum at :$
(a) Point O
(b) Point P
(c) Point K
(d) Point N
›Reveal solutionSolution

For a current-carrying circular coil, the magnetic field on its axis is strongest exactly at the centre and falls off as you move away along the axis.

For a circular coil of radius $R$ carrying current $I$ , the magnetic field at a point on its axis at distance $x$ from the centre is

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

This expression is maximum when $x=0$ , i.e. AT THE CENTRE, giving $B_{max} = \dfrac{\mu_0 I}{2R}$ . As $x$ increases (moving along the axis away from the centre, e.g. towards points P, K, N), the denominator grows and $B(x)$ decreases monotonically. So among O (centre), P, K and N — all of which lie farther from the centre than O — the field is greatest at O.

✓Final answer
$Point O — the field of a circular current loop is maximum at its centre and decreases with distance from it.$
CBSE 2018Set ANNUAL1 markMCQ
$Q. The magnetic field at the centre of a circular coil of radius 5 cm carrying a current of 1 A is 1.256 T. If the radius is made 10 cm, then the magnetic field at the centre of the loop carrying same current is$
(a) 0.628 T
(b) 2.512 T
(c) 5.024 T
(d) 0.314 T
›Reveal solutionSolution

The field at the centre of a circular current loop scales as $1/r$ at fixed current, so doubling the radius halves the field.

Step 1 — Field at centre of a circular loop

$B = \frac{\mu_0 I}{2r} \implies B \propto \frac1r \ \text{(for fixed } I\text{)}$

Step 2 — Scale from $r_1 = 5$ cm to $r_2 = 10$ cm

$B_2 = B_1 \times \frac{r_1}{r_2} = 1.256 \times \frac{5}{10} = 0.628\ \text{T}$

✓Final answer
$B_2 = 0.628$

Q.The value of magnetic field at point O in the given figure is: (A) μ0I2πR\dfrac{\mu_0 I}{2\pi R}2πRμ0​I​ (B) μ0IπR\dfrac{\mu_0 I}{\pi R}πRμ0​I​ (C) μ0I4R\dfrac{\mu_0 I}{4R}4Rμ0​I​ (D) μ0IR\dfrac{\mu_0 I}{R}Rμ0​I​

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Q.The value of magnetic field at point O in the given figure is: (A) $\dfrac{\mu_0 I}{2\pi R}$ (B) $\dfrac{\mu_0 I}{\pi R}$ (C) $\dfrac{\mu_0 I}{4R}$ (D) $\dfrac{\mu_0 I}{R}$

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