A bar magnet of length $l$ and magnetic moment $p_m$ is bent in the form of an arc of radius $r$, subtending an angle of $60°$ at its centre (the two poles of the bent magnet are the ends of the arc, separated by the chord of the $60°$ arc; the total wire length $l$, i.e. the arc length, is unchanged by the bending). (NEET 2013) The new magnetic dipole moment will be (a) $p_m$ (b) $\dfrac{3}{\pi}p_m$ (c) $\dfrac{2}{\pi}p_m$ (d) $\dfrac{1}{2}p_m$

A bar magnet of length l and magnetic moment pm is bent in the form of an arc of radius r, subtending an angle

I. Multiple Choice Questions · Q9

l

(a)

p_m

(b)

\dfrac{3}{\pi}p_m

(c)

\dfrac{2}{\pi}p_m

(d)

\dfrac{1}{2}p_m

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Concept understanding — Magnetic Dipole Moment

Magnetic Dipole Moment – From Intuition to Precision

Think of a bar magnet. It has a north pole and a south pole. If you place it in a magnetic field, it tries to turn — the north pole is pulled one way, the south pole the opposite way. That turning effect (torque) is the most basic sign that something is a magnetic dipole.

A current loop behaves exactly the same way. A circular wire carrying current, when placed in a magnetic field, also feels a torque and tries to align itself. That is the deep insight: a tiny current loop and a bar magnet are the same kind of object — a magnetic dipole.

The Intuitive Picture

Imagine a small, flat loop of wire carrying a steady current $I$ . The loop has an area $A$ . The direction of the loop is defined by its area vector $\mathbf{A}$ — perpendicular to the plane of the loop, following the right-hand rule (curl your fingers along the current, thumb points along $\mathbf{A}$ ).

Now place this loop in a uniform magnetic field $\mathbf{B}$ . What happens?

If the loop is perpendicular to $\mathbf{B}$ , nothing turns — it's already aligned.
If the loop is parallel to $\mathbf{B}$ , it feels maximum torque, trying to flip it perpendicular.
If the loop is at some angle, the torque is somewhere in between.

That torque depends on three things: the current $I$ , the area $A$ , and the angle between the loop and the field. The combination $I A$ is the magnetic dipole moment of the loop.

\mathbf{m} = I \, \mathbf{A}

For a bar magnet, the same idea applies: $\mathbf{m}$ points from the south pole to the north pole (yes, that's the convention — the moment points northward), and its magnitude tells you how strong the dipole is.

The Precise Statement

A magnetic dipole moment $\mathbf{m}$ is a vector that characterises the strength and orientation of a magnetic dipole. For a current loop:

\mathbf{m} = I \, \mathbf{A}

where $I$ is the current and $\mathbf{A}$ is the area vector (magnitude = area, direction = perpendicular to the loop by right-hand rule). For a bar magnet, $\mathbf{m}$ points from south to north, and its magnitude is roughly $m = p \, l$ , where $p$ is the pole strength and $l$ is the separation between poles.

What Happens in a Magnetic Field?

Two key results follow directly from the definition.

Torque: The field tries to align the dipole with itself. The torque is:

\boldsymbol{\tau} = \mathbf{m} \times \mathbf{B}

The magnitude is $\tau = m B \sin\theta$ , where $\theta$ is the angle between $\mathbf{m}$ and $\mathbf{B}$ . Maximum torque when they are perpendicular ( $\theta = 90^\circ$ ), zero when aligned ( $\theta = 0^\circ$ ).

Potential Energy: A dipole in a field has energy that depends on its orientation:

U = -\mathbf{m} \cdot \mathbf{B} = -m B \cos\theta

The lowest energy ( $U = -mB$ ) is when $\mathbf{m}$ is parallel to $\mathbf{B}$ — the stable equilibrium. The highest energy ( $U = +mB$ ) is when they are antiparallel — the unstable equilibrium. …

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Step 1. Originally, $p_m = q_m\,l$ (pole strength times the full straight length $l$ ).

Step 2. After bending into an arc subtending $\theta=60°=\pi/3$ rad, the wire's own length is unchanged (bending doesn't stretch or shorten the wire), so the arc length still equals $l$ : $l = r\theta \Rightarrow r = \dfrac{l}{\theta} = \dfrac{3l}{\pi}$ .

Step 3. The two poles (still carrying the same pole strength $q_m$ , since pole strength does not depend on shape) now sit at the two ends of the arc, separated by the chord, not the arc: …

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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 23 on this concept.

CBSE 2026Set V11 markMCQ
$\vec{m}$
(a) $\theta = 0^\circ$
(b) $\theta = 90^\circ$
(c) $\theta = 180^\circ$
(d) $\theta = 45^\circ$
›Reveal solutionSolution

(a) $\theta = 0^\circ$ …

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CBSE 2026Set A1 markMCQ
$Q. If a bar magnet is cut into two equal pieces transverse to its length, then each piece will (A) lose its magnetism (B) become a single pole (C) behave as a new magnet with a greater magnetic moment (D) behave as a new magnet with reduced magnetic moment$
›Reveal solutionSolution

Cutting a bar magnet across its length gives two shorter magnets, each with a smaller magnetic moment.

Every piece of a magnet is itself a complete magnet with a north and south pole — you can never isolate a single pole. When the bar is cut transverse to its length (across the middle), each half retains the same pole strength $m$ but has half the length $l/2$ . Since magnetic moment is

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Which of the following is a vector quantity?$
(a) Magnetic flux
(b) Magnetic pole strength
(c) Magnetic moment
(d) Permeability
›Reveal solutionSolution

Magnetic (dipole) moment has both magnitude and a fixed direction (from S to N pole, or along the loop's normal by the right-hand rule), making it the only vector among the four options.

Magnetic flux $\Phi_B=\vec{B}\cdot\vec{A}$ is a scalar (a dot product of two vectors gives a scalar). Magnetic pole strength $m$ is defined by the magnitude of pole-to-pole force and is treated as a scalar. Permeability $\mu$ is a scalar material property relating $B$ and $H$ . In contrast, the magnetic moment of a bar magnet or current loop, $\vec{M}=m\,\vec{d}$ (or $\vec{m}=I\vec{A}$ for a current loop), has a definite direction — from the south to the north p …

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CBSE 2026Set ANNUAL1 markMCQ
$B_a = \frac{\mu_0}{4\pi}\frac{2M}{r^3}$
(a) both A and R are true and R is the correct explanation of A.
(b) both A and R are true and R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false.
›Reveal solutionSolution

$B_a=\mu_0(2M)/(4\pi r^3)$ and $B_{eq}=\mu_0M/(4\pi r^3)$ , so indeed $B_a=2B_{eq}$ — both statements are true and R explains A.

The standard bar-magnet field formulas are $B_{axial} = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ and $B_{equatorial} = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$ . Dividing, $\dfrac{B_{axial}}{B_{equatorial}} = \dfrac{2M/r^3}{M/r^3} = 2$ — so the axial field is indeed exactly double the equatorial field at the same distance $r$ , confirming Assertion A is true. Reason R state …

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CBSE 2026Set SEM31 markMCQ
$Q. Magnetic moment of a steel wire is M. If the wire is turned into semicircle by bending it, then the new magnetic moment of the wire will be$
(a) M
(b) 2M/π
(c) M/2π
(d) M/π
›Reveal solutionSolution

Magnetic moment = pole strength × separation. A straight wire has M = m·L; bent into a semicircle of arc length L = πr, the pole separation becomes the diameter 2r, giving M' = m·2r = 2M/π. Option (b).

Step 1 — straight wire: M = m·L, where m is the pole strength and L the length.

Step 2 — bend it into a semicircle. Arc length is preserved: L = πr, so r = L/π.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. When a magnet of magnetic moment m is placed with angle theta in a magnetic field B then the produced potential energy is$
(a) -mB cos theta
(b) -mB sin theta
(c) mB tan theta
(d) Zero
›Reveal solutionSolution

The potential energy of a magnetic dipole in a uniform field is the negative dot product of its moment and the field, giving $-mB\cos\theta$ .

Work must be done against the restoring torque $\tau = mB\sin\theta$ to rotate a magnetic dipole from the reference orientation ( $\theta=90^\circ$ , taken as zero potential energy) to angle $\theta$ :

…

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CBSE 2025Set ANNUAL1 mark
$Q. Write the formula of torque on the magnetic needle of magnetic moment m (vector) when it is allowed to oscillate in the uniform magnetic field B (vector).$
›Reveal solutionSolution

A magnetic needle placed at an angle to a uniform field experiences a torque equal to the cross product of its magnetic moment and the field, which drives its oscillation.

When a magnetic needle of magnetic moment $\vec{m}$ is placed in a uniform magnetic field $\vec{B}$ making angle $\theta$ with it, the field exerts a torque tending to align the needle along $\vec{B}$ :

$\vec{\tau} = \vec{m}\times\vec{B}$ , with magnitude $\tau = mB\sin\theta$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A bar magnet of magnetic moment 'M' is placed in a uniform magnetic field of intensity 'B' at an angle of 'theta' with its direction. The torque applied on it is$
(a) MB
(b) MB cos(theta)
(c) MB(1 - cos(theta))
(d) MB sin(theta)
›Reveal solutionSolution

A magnetic dipole in a uniform field experiences a torque tau = M x B, whose magnitude is MB sin(theta), where theta is the angle between the dipole moment and the field.

For a bar magnet of magnetic moment M placed at angle theta to a uniform field B, the two forces on its poles form a couple. The magnitude of the resulting torque is:

tau = M B sin(theta)

…

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CBSE 2025Set ANNUAL1 markMCQ
$1.20\ Am^2$
(a) $3.6 \times 10^{-5}$ T
(b) $5.8 \times 10^{-5}$ T
(c) $3.6 \times 10^{-4}$ T
(d) $2.2 \times 10^{-4}$ T
›Reveal solutionSolution

The midpoint lies on the equatorial line of both magnets; since both dipole moments point the same way, the two equatorial fields add.

Since the two magnets are placed parallel to each other (side by side) with their north poles pointing the same direction (south), the line joining them is perpendicular to each magnet's own axis — so the midpoint lies on the equatorial line of both magnets, at $r=10\,$ cm $=0.1\,$ m from each.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. A bar magnet of magnetic moment M is cut into two parts of equal lengths. The magnetic moment and pole strength of either part is$
(a) $\frac{M}{2}, \frac{m}{2}$
(b) $M, \frac{m}{2}$
(c) $\frac{M}{2}, m$
(d) $M, m$
›Reveal solutionSolution

Cutting a magnet along its length into two equal halves keeps the pole strength m the same but halves the magnetic moment to M/2 for each piece.

For the original magnet of length $2l$ and pole strength $m$ : $M = m \times 2l$ .

When cut into two equal pieces, each new piece has length $l$ . The pole strength $m$ depends only on the pole face/material and is unaffected by cutting along the length, so it stays $m$ for each piece (a new pole of strength $m$ appears at the cut face, and the old pole of strength $m$ remains …

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CBSE 2024Set A1 markMCQ
$Q. On dividing any magnet of magnetic moment (M) parallel to its length into n equal pieces, the moment of each piece will be (A) M/n (B) M/n^2 (C) M/2n (D) M \times n$
›Reveal solutionSolution
Cutting parallel to the length keeps the length L but divides the pole strength by n, so each moment = M/n.

The magnetic moment of a bar magnet is $M = m\,\ell$ , where m is the pole strength and ℓ its length.

When the magnet is cut into n equal pieces parallel to its length, each piece has:

the same length ℓ (the cuts run along the length), and

a cross-sectional area (and hence pole strength) reduced to $m/n$ , since the pole strength is shared among the n slices.

So each piece has moment

$M' = \left(\frac{m}{n}\right)\ell = \frac{m\ell}{n} = \frac{M}{n}.$

…
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CBSE 2024Set ANNUAL1 markMCQ
$Q. The relation between magnetic field, dipole moment and torque is$
(a) tau = m . B
(b) tau = m x B
(c) tau = m + B
(d) m = tau . B
›Reveal solutionSolution

A magnetic dipole placed in a magnetic field experiences a torque equal to the cross product of its dipole moment and the field, tau = m x B.

A current loop (or a bar magnet) of magnetic dipole moment $\vec{m}$ placed in a uniform magnetic field $\vec{B}$ experiences a torque that tends to align $\vec{m}$ with $\vec{B}$ :

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

…

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