A long straight wire of circular cross-section (radius $a$) carries a steady current $I$. The current is uniformly distributed across this cross-section. The magnitude of the magnetic field produced at a point at a distance $\left(\dfrac{a}{2}\right)$ from the axis of the wire will be (A) Zero (B) $\dfrac{\mu_0 I}{2\pi a}$ (C) $\dfrac{\mu_0 I}{4\pi a}$ (D) $\dfrac{\mu_0 I}{6\pi a}$

> [!TLDR] > 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. 1. **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. 2. **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: $$B \cdot (2\pi r) = \mu_0 I_{\text{enc}}$$ 3. **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: $$I_{\text{enc}} = J \cdot \pi r^2 = \frac{I}{\pi a^2} \cdot \pi r^2 = I \frac{r^2}{a^2}$$ 4. **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 $B$: $$B = \frac{\mu_0 I r}{2\pi a^2}$$ > [!FORMULA] > $$B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2}$$ This is the general expression for the magnetic field at any distance $r \leq a$ from the axis. 5. **Plug in $r = a/2$.** $$B = \f

A long straight wire of circular cross-section (radius a) carries a steady current I. The current is uniformly

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a

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

✓ Free question

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

| Solenoid (ideal) | Rectangle through solenoid | $B = \mu_0 n I$ inside |

| Toroid | Circle inside toroid | $B = \frac{\mu_0 N I}{2\pi r}$ |

What Makes It "Circuital"?

The word "circuital" refers to the closed loop — the circuit you walk around. It has nothing to do with electrical circuits (though the law applies to them too). It's about the circulation of the field.

Think of it this way: magnetic field lines form closed loops. They don't start or end anywhere (unlike electric field lines, which start on positive charges and end on negative ones). Ampere's Law captures this swirling, circulating nature of magnetic fields.

A Common Misconception

Students often think the integral $\oint \vec{B} \cdot d\vec{l}$ depends on the shape of the loop. It doesn't — only on the current enclosed. You can deform the loop any way you like, as long as it still encloses the same current, and the integral gives the same answer.

But the magnetic field $\vec{B}$ at each point on the loop does depend on the shape. The integral magically works out to the same value regardless. This is the power of the law: you choose the loop that makes the calculation easiest.

Important

Ampere's Law is always true, but it's only useful for calculation when the system has enough symmetry to let you pull $B$ out of the integral. For arbitrary current distributions, you need the Biot-Savart law instead.

Ampere's circuital law is a cornerstone CBSE Class 12 Physics NCERT topic, commonly searched as Ampere's circuital law derivation class 12 or Ampere's law applications solenoid toroid. Because it is the fastest route to the solenoid and toroid field formulas tested heavily in JEE Main and NEET physics, mastering when and how to choose the right Amperian loop is a high-value exam skill.

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}

This is the general expression for the magnetic field at any distance $r \leq a$ from the axis.

Plug in $r = a/2$ .

B = \frac{\mu_0 I (a/2)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}

Watch out

A common mistake is to use the formula for the field outside the wire ( $B = \mu_0 I / 2\pi r$ ) for all points. That formula assumes all the current is enclosed, which is only true for $r \geq a$ . Inside the wire, only a fraction of the current contributes, so the field is smaller.

Tip

Notice the linear dependence on $r$ inside the wire. At the surface ( $r = a$ ), the formula gives $B = \mu_0 I / 2\pi a$ , which matches the outside formula at the boundary — a nice consistency check.

✓Final answer

a/2

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 $B$ :
$B = \frac{\mu_0 I r}{2\pi a^2}$ $B_{\text{inside}} = \frac{\mu_0 I r}{2\pi a^2}$
This is the general expression for the magnetic field at any distance $r \leq a$ from the axis.

Plug in $r = a/2$ .

$B = \frac{\mu_0 I (a/2)}{2\pi a^2} = \frac{\mu_0 I}{4\pi a}$
Watch out

A common mistake is to use the formula for the field outside the wire ( $B = \mu_0 I / 2\pi r$ ) for all points. That formula assumes all the current is enclosed, which is only true for $r \geq a$ . Inside the wire, only a fraction of the current contributes, so the field is smaller.

Tip

Notice the linear dependence on $r$ inside the wire. At the surface ( $r = a$ ), the formula gives $B = \mu_0 I / 2\pi a$ , which matches the outside formula at the boundary — a nice consistency check.

✓Final answer
$a/2$
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 $\mu_0$ times the total current threading that path.

✓Final answer
$\oint \vec{B}\cdot d\vec{l}=\mu_0 I$
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

✓Final answer
$(b) at any point inside the pipe is zero$

By Ampere's circuital law, for an Amperian loop inside the hollow pipe the enclosed current is zero (all current flows on the wall). Hence $\oint\vec{B}\cdot d\vec{l}=\mu_0 I_{enc}=0$ , giving $B=0$ at every point inside the pipe.

CBSE 2023Set 55/3/11 markMCQ

r

(a)

(b)

(c)

(d)

›Reveal solution

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
(d)	Hyperbolic $1/r$	No	Correct shape, but no reference to $a$

The field outside the wire is defined for $r \geq a$ , not from $r = 0$ . The vertical dashed line at $r = a$ in graphs (a) and (c) marks the wire's surface, the boundary between inside and outside. Graph (c) combines the correct $1/r$ hyperbolic decay with the correct domain marker.

Watch out

A common mistake is to think the field starts at $r = 0$ . The formula $B = \frac{\mu_0 I}{2\pi r}$ applies only outside the wire. Inside, the field depends on the current distribution and typically grows linearly with $r$ (for a uniform current density). The question asks about the outside region, so the graph must start at $r = a$ .

Graph (d) shows the right functional form but omits the reference to the wire radius $a$ , making it incomplete. Graph (c) is the only one that correctly shows both the $1/r$ hyperbolic fall-off and the starting point at the surface.

✓Final answer

The correct option is (c) .

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).

This law plays the same role for magnetostatics that Gauss's law plays for electrostatics - it is especially useful for finding $\vec{B}$ in situations of high symmetry, e.g. inside a long straight current-carrying wire, a solenoid, or a toroid.

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

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