Verify the Ampere's law for magnetic field of a point dipole of dipole moment mathbfm = mhatk. Take C as the c

NCERT Exemplar · Q19

\mathbf{m} = m\hat{k}

(i) the

z

-axis from

z = a > 0

z = R

;

(ii) along the quarter circle of radius

R

and centre at the origin, in the first quadrant of

x

z

plane;

(iii) along the

x

-axis from

x = R

x = a

, and

(iv) along the quarter circle of radius

a

and centre at the origin in the first quadrant of

x

z

plane.

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Concept understanding — Ampere's Circuital Law

The Intuition: What Ampere's Law Is Really Saying

Imagine you're standing in a field of grass, and you walk in a complete circle. If the grass is perfectly flat and still, your path feels the same all the way around. But if there's a strong wind blowing through the center of your circle, you'll feel it push you differently at different points along your walk.

Electric currents create magnetic fields. Ampere's Circuital Law is a way of measuring how much magnetic field is swirling around a current — like measuring how strong the "whirlpool" of field lines is around a wire.

The key idea: if you take a closed loop (any shape you like) and add up the magnetic field along every tiny piece of that loop, the total you get is directly proportional to the amount of current that passes through the loop. No current through the loop? The total is zero.

Note

This is the magnetic analogue of Gauss's Law for electricity. Gauss's Law relates the flux of electric field through a closed surface to the charge inside. Ampere's Law relates the circulation of magnetic field around a closed loop to the current inside.

The Precise Statement

Ampere's Circuital Law states:

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}

Let's break down every symbol:

$\oint$ — The circle on the integral sign means you're integrating around a closed loop. You start at some point, trace a complete path, and return to where you began.
$\vec{B}$ — The magnetic field at each point on your loop.
$d\vec{l}$ — An infinitesimally small piece of your loop, treated as a vector pointing along the direction you're walking.
$\vec{B} \cdot d\vec{l}$ — The dot product. This picks up only the part of the magnetic field that points along your path. If the field is perpendicular to your path at some point, that piece contributes nothing.
$\mu_0$ — The permeability of free space, a fundamental constant ( $4\pi \times 10^{-7} \, \text{T·m/A}$ ). It tells you how "strongly" a current creates a magnetic field in empty space.
$I_{\text{enc}}$ — The net current passing through the area bounded by your loop. "Net" means you add currents going one way and subtract currents going the opposite way.

Watch out

The current must pass through the loop's opening — not just anywhere near it. A current that runs outside the loop contributes zero to the right-hand side, even if it produces a magnetic field at points on the loop.

Why the Dot Product Matters

The dot product $\vec{B} \cdot d\vec{l} = B \, dl \cos\theta$ where $\theta$ is the angle between the field and your path. This is crucial: if you walk along a path where the magnetic field is always perpendicular to your direction, you get zero contribution at every step — even if the field is huge.

This is why Ampere's Law is most useful for symmetric situations. You choose your loop so that:

The magnetic field is constant in magnitude along the loop.
The field is always parallel (or antiparallel) to your path, so $\cos\theta = \pm 1$ .

Then the integral becomes simple multiplication: $B \times (\text{circumference of loop}) = \mu_0 I_{\text{enc}}$ .

The Classic Example: A Straight Wire

Consider an infinitely long, straight wire carrying current $I$ . The magnetic field circles around the wire in concentric circles. Choose your Amperian loop to be a circle of radius $r$ centered on the wire.

By symmetry, $B$ is the same at every point on the circle and points tangent to it — exactly along $d\vec{l}$ . So:

\oint \vec{B} \cdot d\vec{l} = B \times (2\pi r) = \mu_0 I

Therefore:

B = \frac{\mu_0 I}{2\pi r}

This is the familiar formula for the field around a long straight wire. Notice: the field falls off as $1/r$ , not $1/r^2$ like the electric field from a point charge. Magnetic fields from currents have a different geometry.

Tip

| Configuration | Amperian Loop | Result |

|:---|:---|:---|

| Straight wire | Circle centered on wire | $B = \frac{\mu_0 I}{2\pi r}$ | …

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Ampere's law $\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{enc}}$ is verified for a magnetic dipole by showing the line integral around a closed path that does not enclose the dipole current loop equals zero — the contributions from the four segments cancel pairwise.

Why This Approach Works

The magnetic field of a point dipole $\mathbf{m} = m\hat{k}$ is given by

\mathbf{B}(\mathbf{r}) = \frac{\mu_0}{4\pi} \left[ \frac{3(\mathbf{m}\cdot\hat{r})\hat{r} - \mathbf{m}}{r^3} \right]

This field is not curl-free everywhere — it has a singularity at the origin where the dipole sits. But for any closed loop that does not enclose the dipole (i.e., does not link the current loop that produces it), Ampere's law demands $\oint \mathbf{B} \cdot d\mathbf{l} = 0$ .

The path described in the problem is a closed curve in the $x$ - $z$ plane that stays away from the origin (since $a > 0$ and $R > a$ ). It consists of four segments: two radial lines along the axes and two quarter-circles. The key insight is that along radial lines, $\mathbf{B}$ is either parallel or antiparallel to $d\mathbf{l}$ , while along circular arcs, $\mathbf{B}$ has no tangential component — so the contributions simplify dramatically.

Step-by-Step Verification

1. Set up coordinates and the field components

The dipole moment is $\mathbf{m} = m\hat{k}$ . In spherical coordinates $(r, \theta, \phi)$ , the field is

\mathbf{B} = \frac{\mu_0 m}{4\pi r^3} (2\cos\theta\,\hat{r} + \sin\theta\,\hat{\theta})

In the $x$ - $z$ plane ( $\phi = 0$ ), we have $\hat{r} = \sin\theta\,\hat{x} + \cos\theta\,\hat{z}$ and $\hat{\theta} = \cos\theta\,\hat{x} - \sin\theta\,\hat{z}$ .

Tip

On the $z$ -axis ( $\theta = 0$ or $\pi$ ), $\sin\theta = 0$ so $\mathbf{B}$ is purely radial: $\mathbf{B} = \frac{\mu_0 m}{2\pi r^3} \hat{r}$ . On the $x$ -axis ( $\theta = \pi/2$ ), $\cos\theta = 0$ so $\mathbf{B} = \frac{\mu_0 m}{4\pi r^3} \hat{\theta}$ , and $\hat{\theta} = -\hat{z}$ there — so the field points along $-z$ .

2. Segment (i): along the $z$ -axis from $z = a$ to $z = R$

Here $\theta = 0$ , $r = z$ , and $d\mathbf{l} = dz\,\hat{z} = dr\,\hat{r}$ . The field is

\mathbf{B} = \frac{\mu_0 m}{2\pi r^3} \hat{r}

\mathbf{B} \cdot d\mathbf{l} = \frac{\mu_0 m}{2\pi r^3} dr

Integrating from $r = a$ to $r = R$ :

\int_{(i)} \mathbf{B} \cdot d\mathbf{l} = \frac{\mu_0 m}{2\pi} \int_a^R \frac{dr}{r^3} = \frac{\mu_0 m}{2\pi} \left[ -\frac{1}{2r^2} \right]_a^R = \frac{\mu_0 m}{4\pi} \left( \frac{1}{a^2} - \frac{1}{R^2} \right)

3. Segment (ii): quarter-circle of radius $R$ in first quadrant of $x$ - $z$ plane

On this arc, $r = R$ constant, and $d\mathbf{l} = R\,d\theta\,\hat{\theta}$ (since the path runs clockwise, $\theta$ decreases from $0$ to $\pi/2$ ). The field has both $\hat{r}$ and $\hat{\theta}$ components, but $d\mathbf{l}$ is purely $\hat{\theta}$ , so only the $\hat{\theta}$ component contributes:

\mathbf{B} \cdot d\mathbf{l} = \frac{\mu_0 m}{4\pi R^3} \sin\theta \cdot (R\,d\theta) = \frac{\mu_0 m}{4\pi R^2} \sin\theta\,d\theta

The path goes clockwise: from $\theta = 0$ to $\theta = \pi/2$ . So

\int_{(ii)} \mathbf{B} \cdot d\mathbf{l} = \frac{\mu_0 m}{4\pi R^2} \int_0^{\pi/2} \sin\theta\,d\theta = \frac{\mu_0 m}{4\pi R^2} \left[ -\cos\theta \right]_0^{\pi/2} = \frac{\mu_0 m}{4\pi R^2} (0 + 1) = \frac{\mu_0 m}{4\pi R^2}

Watch out

A common mistake is to forget the sign from the direction of traversal. Here clockwise in the $x$ - $z$ plane means $\theta$ decreases, so $d\theta$ is negative. But we parameterised from $\theta=0$ to $\theta=\pi/2$ and used $d\mathbf{l} = R\,d\theta\,\hat{\theta}$ — the sign is already accounted for by the integration limits. Always check that $d\mathbf{l}$ points along the actual path direction.

4. Segment (iii): along the $x$ -axis from $x = R$ to $x = a$

On the $x$ -axis, $\theta = \pi/2$ , $r = x$ , and $d\mathbf{l} = dx\,\hat{x}$ . The field is …

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Method: Verifying Ampere's Law ( $\oint\vec B\cdot d\vec l = \mu_0 I_{\text{enc}}$ ) Along a Given Multi-Segment Path

Use this general technique whenever you're asked to explicitly verify Ampere's law by computing the line integral of a known field around a specified closed path made of several distinct segments (straight radial pieces and circular arcs).

Steps

Step 1: Write the field in spherical components matched to the path's geometry

For a point dipole $\vec m=m\hat k$ , the field in spherical coordinates is

\vec B = \frac{\mu_0 m}{4\pi r^3}\left(2\cos\theta\,\hat r + \sin\theta\,\hat\theta\right).

Having both a radial ( $\hat r$ ) and a polar ( $\hat\theta$ ) piece explicitly separated is what makes the path integral tractable — each type of path segment will only "see" one of the two components.

Step 2: Classify every segment of the path by which unit vector its $d\vec l$ points along

A segment along a radial line (fixed $\theta$ , varying $r$ ) has $d\vec l = dr\,\hat r$ , so only the $\hat r$ component of $\vec B$ contributes: $\vec B\cdot d\vec l = B_r\,dr$ .
A segment along a circular arc (fixed $r$ , varying $\theta$ ) has $d\vec l = r\,d\theta\,\hat\theta$ , so only the $\hat\theta$ component contributes: $\vec B\cdot d\vec l = B_\theta\,r\,d\theta$ .
A segment lying on an axis where the field happens to be purely along the other unit vector (e.g. on the axis perpendicular to $\vec m$ , where $B_r=0$ ) contributes zero, since $\vec B$ there is perpendicular to $d\vec l$ — always check for this shortcut before integrating blindly.

Step 3: Integrate each segment in turn, keeping track of the direction of traversal …

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

CBSE 2026Set 55/1/11 markMCQ
$a$
›Reveal solutionSolution
For a current uniformly distributed across a wire's cross-section, the magnetic field inside the wire grows linearly with distance from the axis. At $r = a/2$ , the field is half its surface value, giving $\boxed{\dfrac{\mu_0 I}{4\pi a}}$ , which corresponds to option (C).

The key insight here is that the magnetic field inside a current-carrying conductor depends only on the current enclosed by the Amperian loop, not the total current. For a uniform current density, the enclosed current scales with the area of the loop, so the field inside increases linearly with $r$ .

Let's work through this systematically.

Set up the problem. We have a long straight wire of radius $a$ carrying a steady current $I$ , uniformly distributed over its cross-section. We need the magnetic field at a distance $r = a/2$ from the axis — that's a point inside the wire.

Recall Ampere's Law. For a long straight wire with cylindrical symmetry, Ampere's Law states:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}$
where $I_{\text{enc}}$ is the current passing through the surface bounded by the Amperian loop. By symmetry, $\vec{B}$ is tangential and constant in magnitude along a circular path of radius $r$ , so:
$B \cdot (2\pi r) = \mu_0 I_{\text{enc}}$

Find the current enclosed at $r = a/2$ . Since the current is uniform, the current density is:

$J = \frac{I}{\pi a^2}$
The area enclosed by our Amperian loop of radius $r$ is $\pi r^2$ , so:
$I_{\text{enc}} = J \cdot \pi r^2 = \frac{I}{\pi a^2} \cdot \pi r^2 = I \frac{r^2}{a^2}$

Apply Ampere's Law. Substitute $I_{\text{enc}}$ into the equation:

$B \cdot (2\pi r) = \mu_0 \left( I \frac{r^2}{a^2} \right)$
Solve for $B$ :
$B = \frac{\mu_0 I r}{2\pi a^2}$ $B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2}$
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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following formula represents Ampere's circuital law?$
(i) $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$
(ii) $\varphi_E = \dfrac{1}{\epsilon_0}(q)$
(iii) $d\vec{B} = \dfrac{\mu_0}{4\pi}\dfrac{I\,d\vec{l}\times\vec{r}}{r^3}$
(iv) $I = \dfrac{V}{R}$
›Reveal solutionSolution

Ampere's circuital law relates the line integral of B around a closed loop to the enclosed current.

Ampere's circuital law states that the line integral of magnetic field $\vec{B}$ around any closed path equals $\mu_0$ t …

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CBSE 2024Set A11 markMCQ
$Q. A current I flows along the length of an infinitely long, straight thin walled pipe, then the magnetic field$
(a) at all points inside the pipe is same but not zero
(b) at any point inside the pipe is zero
(c) is zero only on the axis of the pipe
(d) is different at different points inside the pipe
›Reveal solutionSolution

(b) at any point inside the pipe is zero …

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CBSE 2023Set 55/3/11 markMCQ
$r$
(a)
(b)
(c)
(d)
›Reveal solutionSolution
Figure — CBSE 2023 55/3/1 Q4

Outside an infinite current-carrying wire, Ampère's law gives $B = \frac{\mu_0 I}{2\pi r}$ , a $1/r$ hyperbola starting at the surface $r = a$ . The correct graph is (c).

The magnetic field around a long straight wire is one of the cleanest applications of Ampère's circuital law. The key insight is that the field depends only on the current enclosed by your Amperian loop, and symmetry forces the field to be tangent to circles centered on the wire.

For a wire of radius $a$ carrying current $I$ , the field behaves differently inside and outside. We care about the outside region, $r \geq a$ .

Why the field varies as $1/r$

Ampère's law on a circular loop of radius $r > a$ Draw a circle of radius $r$ centered on the wire. By symmetry, $\vec{B}$ is constant in magnitude along this circle and tangent to it. Ampère's law states:

$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enc}}.$
The left side is simply $B \cdot 2\pi r$ (the field magnitude times the circumference). The enclosed current is the total wire current $I$ . So:
$B \cdot 2\pi r = \mu_0 I.$

Solve for $B$ Rearranging:

$B = \frac{\mu_0 I}{2\pi r}.$
This is an inverse relationship: as you move farther from the wire, the field drops off as $1/r$ . Mathematically, this is a rectangular hyperbola.

The domain: $r \geq a$ The formula $B = \frac{\mu_0 I}{2\pi r}$ applies outside the wire, meaning $r \geq a$ . At the surface $r = a$ , the field reaches its maximum value for the outside region:

$B_{\text{max}} = \frac{\mu_0 I}{2\pi a}.$
For $r > a$ , the field decreases smoothly as $1/r$ .
$B(r) = \frac{\mu_0 I}{2\pi r}, \quad r \geq a.$

Reading the graphs

Now compare the four options:

Graph Shape Surface marker at $r=a$ Verdict
(a) Linear fall-off Yes Wrong shape (not $1/r$ )
(b) Linear fall-off No Wrong shape, no reference to $a$
(c) Hyperbolic $1/r$ Yes Correct
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CBSE 2017Set ANNUAL1 mark
$Q. State Ampere's circuital law.$
›Reveal solutionSolution

The circulation of $\vec{B}$ around a closed path is $\mu_0$ times the current threading that path.

Ampere's circuital law relates the magnetic field around a closed loop to the electric current passing through the loop. For any closed curve (Amperian loop),

$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{enc}$

where $I_{enc}$ is the net steady current enclosed by the loop and $\mu_0 = 4\pi\times10^{-7}\,\text{T m/A}$ is the permeability of free space. Only currents that actually pass through the surface bounded by the loop contribute to $I_{enc}$ ; currents outside the loop don't contribute to the line integral (though they can still affect $\vec{B}$ locally, their net contribution to the loop integral cancels).

…

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Graph	Shape	Surface marker at $r=a$	Verdict
(a)	Linear fall-off	Yes	Wrong shape (not $1/r$ )
(b)	Linear fall-off	No	Wrong shape, no reference to $a$
(c)	Hyperbolic $1/r$	Yes	Correct

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Q.Verify the Ampere's law for magnetic field of a point dipole of dipole moment m=mk^\mathbf{m} = m\hat{k}m=mk^. Take C as the closed curve running clockwise along

Why This Approach Works

Step-by-Step Verification

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Q.Verify the Ampere's law for magnetic field of a point dipole of dipole moment $\mathbf{m} = m\hat{k}$ . Take C as the closed curve running clockwise along

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