A wire of length $4{\cdot}4$ m is bent round in the shape of a circular loop and carries a current of $1{\cdot}0$ A. The magnetic moment of the loop will be : (A) $0{\cdot}7\ \text{Am}^2$ (B) $1{\cdot}54\ \text{Am}^2$ (C) $2{\cdot}10\ \text{Am}^2$ (D) $3{\cdot}5\ \text{Am}^2$

A wire of length 4·4 m is bent round in the shape of a circular loop and carries a current of 1·0 A. The magne

Question

4{\cdot}4

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

✓ Free question

Concept understanding — Magnetic Moment of a Current Loop

Magnetic Moment of a Current Loop

Think of a tiny compass needle. It has a north pole and a south pole, and when you put it in a magnetic field, it feels a torque — a twist that tries to align it with the field. A current-carrying loop behaves exactly the same way. It's not a permanent magnet, but it acts like one: it has its own magnetic "strength" and a preferred direction, and an external field will try to rotate it.

That "magnetic strength" of the loop is called its magnetic moment. The bigger the current, the stronger the effect. The larger the area enclosed by the loop, the stronger the effect. And if you stack many loops together (N turns), each turn contributes, so the effect multiplies.

The Precise Definition

For a flat loop of wire carrying a steady current, the magnetic moment $\vec{m}$ is defined as:

\vec{m} = N I \vec{A}

Here:

$N$ is the number of turns of wire (if it's a single loop, $N=1$ ).
$I$ is the current flowing through the loop.
$\vec{A}$ is a vector whose magnitude equals the area enclosed by the loop, and whose direction is perpendicular to the plane of the loop.

The direction of $\vec{A}$ — and therefore of $\vec{m}$ — is given by the right-hand rule: curl the fingers of your right hand in the direction of the current; your thumb points along the magnetic moment.

Important

The magnetic moment is a vector quantity. Its magnitude is $m = NIA$ , and its direction is normal to the plane of the loop, following the right-hand rule.

Why This Makes Sense

A single moving charge produces a magnetic field. A current is a stream of moving charges. When those charges go around a loop, their individual magnetic fields add up. The net effect far away from the loop is identical to that of a tiny bar magnet placed at the centre of the loop, with its north pole pointing along $\vec{m}$ .

The torque this loop experiences in a uniform external magnetic field $\vec{B}$ is:

\vec{\tau} = \vec{m} \times \vec{B}

The magnitude of the torque is $mB\sin\theta$ , where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$ . This is exactly the same formula as for a bar magnet. The loop tries to align its magnetic moment with the field — just like a compass needle.

Tip

For a square loop of side $L$ , area $A = L^2$ . For a circular loop of radius $r$ , area $A = \pi r^2$ . The formula $m = NIA$ works for any flat shape.

A Common Exam Point

If the loop is not flat — say it's a solenoid — the same idea applies. A solenoid is just many circular loops stacked together. Its magnetic moment is still $NIA$ , where $A$ is the cross-sectional area of the solenoid and $N$ is the total number of turns. The direction is along the axis of the solenoid, again by the right-hand rule.

Watch out

Do not confuse magnetic moment with magnetic field. The magnetic moment is a property of the source (the loop). The magnetic field is the effect it produces in the space around it. They are related but not the same thing.

The Bottom Line

Magnetic moment of a current loop: $m = NIA$ , directed perpendicular to the plane of the loop by the right-hand rule. It quantifies how strongly the loop behaves like a magnet and determines the torque it experiences in an external field.

The magnetic moment of a current loop is a formula students often search for as magnetic moment of a current loop formula class 12 or magnetic moment and torque important questions. This NCERT Class 12 Physics concept feeds directly into torque and galvanometer problems tested in CBSE boards, JEE Main, and NEET physics alike.

The magnetic moment of a current loop is $M = NIA$ , where $N=1$ , $I=1.0\ \text{A}$ , and the area $A$ comes from the wire length $L = 4.4\ \text{m}$ forming the circumference. The radius is $r = L/(2\pi) \approx 0.700\ \text{m}$ , so $A = \pi r^2 \approx 1.54\ \text{m}^2$ , giving $M \approx 1.54\ \text{Am}^2$ . The correct option is (B).

Concept and intuition

The magnetic moment of a current-carrying loop is a measure of how strong a magnet the loop behaves like. For a single turn, it’s simply the product of the current and the area enclosed: $M = I A$ . The wire length here is the circumference of the loop — that’s the only link between the given length and the area. So the problem reduces to: given the circumference, find the radius, then the area, then multiply by current.

A common mistake is to forget that the wire length is the circumference, not the diameter or something else. Also, watch the decimal: $4.4$ m is exact, and the answer choices are given to two decimal places, so we need to compute carefully.

Step-by-step solution

Identify the loop geometry

The wire is bent into a single circular loop, so the number of turns $N = 1$ . The total length of the wire $L = 4.4\ \text{m}$ is exactly the circumference of the circle.
Find the radius from the circumference

Circumference $C = 2\pi r = L$ .

r = \frac{L}{2\pi} = \frac{4.4}{2\pi} = \frac{2.2}{\pi}

Using $\pi \approx 3.1416$ ,

r \approx \frac{2.2}{3.1416} \approx 0.700\ \text{m}

(You can keep it as $\frac{2.2}{\pi}$ for exactness until the final step.)

Compute the area of the loop Area of a circle: $A = \pi r^2$ .

A = \pi \left( \frac{2.2}{\pi} \right)^2 = \pi \cdot \frac{4.84}{\pi^2} = \frac{4.84}{\pi}

Numerically,

A \approx \frac{4.84}{3.1416} \approx 1.540\ \text{m}^2

Calculate the magnetic moment For a single-turn loop, $M = I A$ .

M = 1.0\ \text{A} \times 1.540\ \text{m}^2 = 1.54\ \text{Am}^2

Tip

You can avoid computing $r$ separately: $A = \frac{L^2}{4\pi}$ directly from $A = \pi (L/(2\pi))^2$ . Here $L=4.4$ , so $A = \frac{4.4^2}{4\pi} = \frac{19.36}{4\pi} = \frac{4.84}{\pi}$ , same result.

Match with the options The value $1.54\ \text{Am}^2$ corresponds exactly to option (B).

Watch out

If you mistakenly used $L$ as the diameter, you’d get $r=2.2$ m, $A \approx 15.2\ \text{m}^2$ , and $M \approx 15.2\ \text{Am}^2$ — not among the options. Always check: a wire bent into a circle means its length is the circumference.

✓Final answer

1.54\ \text{Am}^2

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 A1 markMCQ
$Q. A circular loop radius R carries a current I. Its magnetic dipole moment is (A) IR^2 (B) IR (C) 2π IR (D) πIR^2$
›Reveal solutionSolution

Magnetic dipole moment m = current × area = I·πR² for a circular loop of radius R.

The magnetic dipole moment of a plane current loop is

$m = N I A$

where $N$ is the number of turns (here 1), $I$ the current, and $A$ the area enclosed. For a circular loop of radius $R$ , the area is $A = \pi R^2$ , so

$m = I \times \pi R^2 = \pi I R^2.$

✓Final answer
$(D) πIR².$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: If coil is viewed from one end and the current flows in clockwise direction, then this end is a ............... pole.$
›Reveal solutionSolution

By the clock rule for a current-carrying coil: clockwise current (as viewed) marks a South pole, anticlockwise marks a North pole.

For a current loop behaving as a magnet, use the right-hand rule (curl fingers along the current direction, thumb points to the North pole face). Equivalently, using the simple clock rule: if current appears to flow clockwise when viewed from a given end, that end behaves as a South pole; if it flows anticlockwise, that end is a North pole.

✓Final answer
$The end is a South pole.$
CBSE 2025Set 55/5/11 markMCQ
$\dfrac{10}{\pi}$
›Reveal solutionSolution
The magnetic dipole moment of a current loop equals the product of current and loop area; for a circular loop with $I = \frac{10}{\pi}$ A and $r = 0.14$ m, we get $M = 0.196$ A m².

The magnetic dipole moment is a measure of how strongly a current loop behaves as a magnetic source. Just as an electric dipole has a dipole moment $p = qd$ , a current loop creates a magnetic field pattern identical to that of a bar magnet, characterized by its magnetic moment.

The key insight is that the magnetic dipole moment depends on two factors: how much current flows (the strength of the moving charges) and how large an area the current encloses (the spatial extent of the loop). The larger either quantity, the stronger the magnetic effect.
$M = I \cdot A$
where $M$ is the magnetic dipole moment, $I$ is the current, and $A$ is the area enclosed by the loop.

Now let's calculate step by step:

Identify the given quantities

Current: $I = \frac{10}{\pi}$ A

Radius: $r = 14$ cm $= 0.14$ m

Calculate the area of the circular loop

For a circle, $A = \pi r^2$ :

$A = \pi \times (0.14)^2 = \pi \times 0.0196 \text{ m}^2$

Compute the magnetic dipole moment Substituting into the formula:

$M = I \times A = \frac{10}{\pi} \times \pi \times 0.0196$
The $\pi$ terms cancel:
$M = 10 \times 0.0196 = 0.196 \text{ A m}^2$
Tip

Notice how the problem gives current as $\frac{10}{\pi}$ and the radius such that $\pi r^2$ contains $\pi$ — this is designed so the $\pi$ factors cancel cleanly. Always look for such simplifications before reaching for a calculator.

✓Final answer
$\boxed{0.196 \text{ A m}^2}$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Two wires of same length are shaped into a square and a circle. If they carry the same current, the ratio of the magnetic moments is$
(i) 2 : pi
(ii) pi : 2
(iii) pi : 4
(iv) 4 : pi
›Reveal solutionSolution

Magnetic moment m = IA; comparing areas for equal perimeter L gives square : circle = pi : 4.

For the square, side $= L/4$ , area $A_s = (L/4)^2 = L^2/16$ . For the circle, circumference $L = 2\pi r \Rightarrow r = L/2\pi$ , area $A_c = \pi r^2 = \pi \cdot \frac{L^2}{4\pi^2} = \frac{L^2}{4\pi}$ . Same current I, so $m = IA$ and

$\frac{m_s}{m_c} = \frac{A_s}{A_c} = \frac{L^2/16}{L^2/4\pi} = \frac{4\pi}{16} = \frac{\pi}{4}.$

Hence the ratio is pi : 4.

✓Final answer
$(iii) pi : 4.$
CBSE 2024Set 55/2/11 markMCQ
$4{\cdot}4$
›Reveal solutionSolution
The magnetic moment of a current loop is $M = NIA$ , where $N=1$ , $I=1.0\ \text{A}$ , and the area $A$ comes from the wire length $L = 4.4\ \text{m}$ forming the circumference. The radius is $r = L/(2\pi) \approx 0.700\ \text{m}$ , so $A = \pi r^2 \approx 1.54\ \text{m}^2$ , giving $M \approx 1.54\ \text{Am}^2$ . The correct option is (B).

Concept and intuition

The magnetic moment of a current-carrying loop is a measure of how strong a magnet the loop behaves like. For a single turn, it’s simply the product of the current and the area enclosed: $M = I A$ . The wire length here is the circumference of the loop — that’s the only link between the given length and the area. So the problem reduces to: given the circumference, find the radius, then the area, then multiply by current.

A common mistake is to forget that the wire length is the circumference, not the diameter or something else. Also, watch the decimal: $4.4$ m is exact, and the answer choices are given to two decimal places, so we need to compute carefully.

Step-by-step solution

Identify the loop geometry

The wire is bent into a single circular loop, so the number of turns $N = 1$ . The total length of the wire $L = 4.4\ \text{m}$ is exactly the circumference of the circle.

Find the radius from the circumference

Circumference $C = 2\pi r = L$ .

$r = \frac{L}{2\pi} = \frac{4.4}{2\pi} = \frac{2.2}{\pi}$
Using $\pi \approx 3.1416$ ,
$r \approx \frac{2.2}{3.1416} \approx 0.700\ \text{m}$
(You can keep it as $\frac{2.2}{\pi}$ for exactness until the final step.)

Compute the area of the loop Area of a circle: $A = \pi r^2$ .

$A = \pi \left( \frac{2.2}{\pi} \right)^2 = \pi \cdot \frac{4.84}{\pi^2} = \frac{4.84}{\pi}$
Numerically,
$A \approx \frac{4.84}{3.1416} \approx 1.540\ \text{m}^2$

Calculate the magnetic moment For a single-turn loop, $M = I A$ .

$M = 1.0\ \text{A} \times 1.540\ \text{m}^2 = 1.54\ \text{Am}^2$
Tip

You can avoid computing $r$ separately: $A = \frac{L^2}{4\pi}$ directly from $A = \pi (L/(2\pi))^2$ . Here $L=4.4$ , so $A = \frac{4.4^2}{4\pi} = \frac{19.36}{4\pi} = \frac{4.84}{\pi}$ , same result.

Match with the options The value $1.54\ \text{Am}^2$ corresponds exactly to option (B).

Watch out

If you mistakenly used $L$ as the diameter, you’d get $r=2.2$ m, $A \approx 15.2\ \text{m}^2$ , and $M \approx 15.2\ \text{Am}^2$ — not among the options. Always check: a wire bent into a circle means its length is the circumference.

✓Final answer
$1.54\ \text{Am}^2$
CBSE 2023Set 55/3/11 markMCQ
$Q. Two statements are given - one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer from the codes (a), (b),$
(c) and
(d) below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is also false. Assertion (A) : When the radius of a circular loop carrying a steady current is doubled, its magnetic moment becomes four times. Reason (R) : The magnetic moment of a circular loop carrying a steady current is proportional to the area of the loop.
›Reveal solutionSolution
The magnetic moment of a current loop is $M = I \cdot A$ , so doubling the radius quadruples the area and thus the magnetic moment. Both Assertion and Reason are true, and the Reason correctly explains the Assertion — answer is (a).

The heart of this question is the magnetic moment of a current-carrying loop. Magnetic moment is the fundamental quantity that determines how a current loop behaves in an external magnetic field — it experiences a torque trying to align it with the field, just like a compass needle.

For any planar loop carrying a steady current $I$ , the magnetic moment $\vec{M}$ is defined as:
$\vec{M} = I \cdot \vec{A}$
where $\vec{A}$ is the area vector of the loop (magnitude = area, direction = perpendicular to the plane, given by the right-hand rule). This is a direct, proportional relationship — double the current, double the moment; double the area, double the moment.

Now let's apply this to the specific statements.

Assertion (A): A circular loop of radius $r$ has area $A = \pi r^2$ . Its magnetic moment is $M = I \cdot \pi r^2$ . If the radius is doubled to $2r$ , the new area becomes $A' = \pi (2r)^2 = 4\pi r^2 = 4A$ . Since current $I$ is unchanged (steady current), the new magnetic moment is $M' = I \cdot 4\pi r^2 = 4M$ . So the magnetic moment indeed becomes four times. Assertion (A) is true.

Reason (R): The statement says magnetic moment is proportional to the area of the loop. From $M = I A$ , with $I$ constant, $M \propto A$ . This is exactly correct. Reason (R) is true.

Does Reason correctly explain Assertion? The Assertion's result (fourfold increase) follows directly from the proportionality stated in the Reason. When radius doubles, area quadruples, and because $M \propto A$ , the magnetic moment quadruples. The Reason is the precise physical law that produces the Assertion's outcome.

Watch out

A common mistake is to think magnetic moment depends on radius directly (like $M \propto r$ ) rather than on area ( $M \propto r^2$ ). Always go back to $M = I A$ — the area is what matters, not the circumference or diameter.

Tip

For a quick check: if the loop were a square of side $L$ , area $L^2$ , doubling the side quadruples the area and the magnetic moment. The same logic holds for any shape — magnetic moment is always $I \times$ area for a planar loop.

✓Final answer
$The correct option is (a) : Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).$
CBSE 2023Set F1 markMCQ
$Q. The radius of a circular current loop is made double and the current is made half. The magnetic moment of the loop will become (A) Halved (B) Doubled (C) Four times large (D) None of these$
›Reveal solutionSolution

$M = I\pi r^2$ ; area ×4 and current ×½ ⇒ moment ×2.

The magnetic moment of a single circular current loop is

$M = I A = I\,\pi r^2.$

New radius $r' = 2r \Rightarrow$ area becomes $\pi (2r)^2 = 4\pi r^2$ (×4).

New current $I' = I/2$ (×½).

So $M' = \left(\dfrac{I}{2}\right)\pi(2r)^2 = \dfrac{I}{2}\cdot 4\pi r^2 = 2\,I\pi r^2 = 2M.$

The magnetic moment doubles.

✓Final answer
$(B) Doubled.$
CBSE 2023Set ANNUAL1 mark
$Q. Current loop behaves as a ________. (Magnetic dipole/Electric dipole)$
›Reveal solutionSolution

A current-carrying loop produces a magnetic field pattern identical to that of a bar magnet (a magnetic dipole), with a magnetic moment $m = NIA$ .

A planar loop of area A carrying current I has a magnetic moment $\vec{m} = I\vec{A}$ (or $NIA$ for N turns), directed normal to the plane of the loop as given by the right-hand rule. Far from the loop, its magnetic field falls off with distance and is oriented exactly like the field of a short bar magnet with a north and south pole. This loop–magnet equivalence is why a current loop is called a magnetic dipole, analogous to how two opposite charges separated by a small distance form an electric dipole.

✓Final answer
$A current loop behaves as a magnetic dipole (not an electric dipole).$
CBSE 2023Set ANNUAL1 markMCQ
$n$
(a) $n^2IA$
(b) $nIA^2$
(c) $nIA$
(d) $nI^2A$
›Reveal solutionSolution

The magnetic (dipole) moment of a current loop is the product of the number of turns, the current and the area.

Solution:

For a plane current loop of $n$ turns, each carrying current $I$ and enclosing area $A$ , each turn behaves as a magnetic dipole of moment $IA$ (current × area), directed normal to the plane of the loop (right-hand rule). Since the $n$ turns are effectively $n$ such dipoles with the same orientation, their moments add directly:

$m = n \times (IA) = nIA$

✓Final answer
$nIA$
CBSE 2022Set I1 markMCQ
$Q. Electromagnetic moment of current carrying coil is (A) NI A (B) N A/I (C) N/(I A) (D) I A/N$
›Reveal solutionSolution

The magnetic (dipole) moment of a coil is m = N I A.

For a plane coil of N turns, each carrying current I and enclosing area A, the magnetic dipole moment is

$m = N I A$

directed along the coil axis (right-hand rule). Its SI unit is A·m². The other options have wrong dimensions.

✓Final answer
$(A) NI A.$
CBSE 2022Set I1 markMCQ
$Q. Electromagnetic moment of current carrying coil is (A) m = N A/I (B) m = A/(NI) (C) m = NI A (D) m = I A/N$
›Reveal solutionSolution

Magnetic moment of a coil $m = NIA$ .

A single planar loop carrying current $I$ and enclosing area $A$ has a magnetic dipole moment $m = IA$ , directed normal to the plane of the loop (right-hand rule).

If the coil has $N$ identical turns, each turn contributes $IA$ , so the total magnetic moment is

$m = NIA.$

Its SI unit is A·m². The other options have incorrect dependence on $N$ or $I$ .

✓Final answer
$(C) m = NI A.$
CBSE 2020Set 55/1/11 markMCQ
$Q. The magnetic dipole moment of a current carrying coil does not depend upon (A) number of turns of the coil. (B) cross-sectional area of the coil. (C) current flowing in the coil. (D) material of the turns of the coil.$
›Reveal solutionSolution
The magnetic dipole moment $\mu = NIA$ depends only on the current, number of turns, and loop area — not on what the wire is made of. The answer is (D).

The magnetic dipole moment is a measure of the strength of a current loop as a magnetic source. To understand what it depends on, we need to see where it comes from.

When a steady current flows through a closed loop, it creates a magnetic field pattern identical to that of a tiny bar magnet. The strength of this "equivalent magnet" is quantified by the magnetic dipole moment. For a single circular loop carrying current $I$ and enclosing area $A$ , the dipole moment is:
$\mu = IA$
This is a vector quantity, with direction given by the right-hand rule (curl your fingers along the current, thumb points along $\mu$ ).

Now let's examine each factor systematically:

Number of turns ( $N$ ): If you stack $N$ identical loops carrying the same current, each contributes $IA$ to the total moment. They add up because the magnetic fields from each turn reinforce one another. The total moment becomes:

$\mu = NIA$
So the dipole moment is directly proportional to the number of turns.

Cross-sectional area ( $A$ ): A larger loop encloses more area, which means more magnetic flux threads through it and a stronger field is produced at distant points. The moment scales linearly with area — double the area, double the moment.

Current ( $I$ ): The magnetic field strength is directly proportional to the current flowing. More current means more moving charge per unit time, hence a stronger magnetic effect. The moment is directly proportional to $I$ .

Material of the wire: Here's the key insight. The magnetic dipole moment depends only on the current (charge flow rate), not on how that current is achieved. Whether the wire is copper, aluminum, silver, or any other conductor, as long as the same current $I$ flows through the same geometry, the magnetic moment is identical.

Note

The material affects the resistance of the coil (and hence how much voltage you need to drive a given current), but once the current is established, the magnetic moment is determined purely by $I$ , $N$ , and $A$ .

The formula $\mu = NIA$ contains no material-dependent parameters like resistivity, permeability of the conductor, or electron density. The magnetic field arises from moving charges, and the dipole moment counts only the total current and the area it encircles.

Watch out

Don't confuse the material of the wire with the material of the core. If you insert a ferromagnetic core inside the coil, that dramatically changes the effective magnetic moment because the core magnetizes and adds its own contribution. But the question asks about the material of the turns (the wire itself), which is irrelevant to $\mu$ .

✓Final answer
$The correct option is (D) — the magnetic dipole moment does not depend on the material of the turns of the coil.$

Concept and intuition

Step-by-step solution

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