Q.A closely wound circular coil of $50$ turns and area $2\times10^{-2}\ \text{m}^2$ carries a current of $2\ \text{A}$. Calculate the magnitude of its magnetic dipole moment.

Showing the 12 most recent of 23 on this concept.

1. CBSE 2026Set V11 markMCQ
   Q.A magnetic dipole of magnetic moment $\vec{m}$ is placed in a uniform magnetic field $\vec{B}$ such that the angle between $\vec{m}$ and $\vec{B}$ is $\theta$. If the magnetic dipole is in stable equilibrium position, then :

   (a) $\theta = 0^\circ$

   (b) $\theta = 90^\circ$

   (c) $\theta = 180^\circ$

   (d) $\theta = 45^\circ$

   ›Reveal solutionSolution

   (a) $\theta = 0^\circ$

   ✓Final answer

   (a) $\theta = 0^\circ$

   The potential energy of a magnetic dipole in a field is $U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$. This is minimum when $\cos\theta$ is maximum, i.e. $\theta = 0^\circ$ (dipole aligned with the field). Minimum potential energy corresponds to stable equilibrium; $\theta = 180^\circ$ is unstable equilibrium.

2. 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

   $\text{moment}=m\times l,$

   halving the length halves the moment. So each piece is a new magnet with a reduced magnetic moment.

   ✓Final answer

   (D) behave as a new magnet with reduced magnetic moment.

3. 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 pole of the magnet, or along the loop's area-vector normal given by the right-hand rule — so it must be added/combined vectorially, e.g. when a magnet is placed in an external field, torque $\vec{\tau}=\vec{m}\times\vec{B}$ depends on this direction.

   ✓Final answer

   (c) Magnetic moment

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Assertion (A): The magnetic field strength due to a bar magnet on its axial line at a distance r is double than at the equatorial line at the same distance. Reason (R): Magnetic field strength at a point on the axial line is $B_a = \frac{\mu_0}{4\pi}\frac{2M}{r^3}$ and on the equatorial line is $B_{eq} = \frac{\mu_0}{4\pi}\frac{M}{r^3}$.

   (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 states these two formulas correctly, and they are precisely what was just used to derive the factor of 2 in A — so R correctly explains A.

   ✓Final answer

   (a) Both A and R are true, and R is the correct explanation of A.

5. 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.

   Step 3 — the two poles are now at the ends of the semicircle, separated by the diameter 2r. Pole strength m is unchanged.

   M' = m·(2r) = m·(2L/π) = (2/π)(mL) = 2M/π.

   This reshaping-of-a-magnet result belongs to NCERT/CBSE Class 12 Physics, Magnetism and Matter.

   ✓Final answer

   (b) 2M/π

6. 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 $\theta$:

   $U(\theta) = \int_{90^\circ}^{\theta} \tau\, d\theta' = \int_{90^\circ}^{\theta} mB\sin\theta'\,d\theta' = -mB\cos\theta$

   Equivalently, $U = -\vec{m}\cdot\vec{B} = -mB\cos\theta$.

   ✓Final answer

   (a) -mB cos theta.

7. 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 $\tau = mB\sin\theta$

   This restoring torque is what makes a displaced needle oscillate (like a physical pendulum) about its equilibrium orientation, forming the basis of the oscillation (vibration) magnetometer.

   ✓Final answer

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

8. 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)

   This torque is zero when theta = 0 (aligned, stable equilibrium) or theta = 180 degrees (anti-aligned, unstable equilibrium), and maximum when theta = 90 degrees - consistent with the cross-product form tau = M x B.

   ✓Final answer

   (d) MB sin(theta).

9. CBSE 2025Set ANNUAL1 markMCQ
   Q.Two short bar magnets of 1cm have magnetic moments $1.20\ Am^2$ and $1.00\ Am^2$ respectively. They are placed on a horizontal table parallel to each other and north pole pointing towards south. What is the magnitude of resultant magnetic field at the middle of line joining them if their separation is 20cm? (neglect earth's magnetic field)

   (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.

   Equatorial field of a short bar magnet: $B_{eq} = \dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$, and this field points antiparallel to the magnet's own dipole moment. Since both magnets' moments point the same way, both equatorial fields at the midpoint also point the same way, so they add:

   $B_1 = 10^{-7}\times\frac{1.20}{(0.1)^3} = 10^{-7}\times1200 = 1.2\times10^{-4}\ \text{T}$

   $B_2 = 10^{-7}\times\frac{1.00}{(0.1)^3} = 10^{-7}\times1000 = 1.0\times10^{-4}\ \text{T}$

   $B_{net} = B_1+B_2 = 2.2\times10^{-4}\ \text{T}$

   ✓Final answer

   (d) $2.2\times10^{-4}$ T.

10. 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 at the other end).

    New moment of each piece: $M' = m \times l = \frac{m \times 2l}{2} = \frac{M}{2}$

    So each half has magnetic moment $M/2$ and pole strength $m$ (unchanged).

    ✓Final answer

    The correct option is (c) M/2, m.

11. 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 × 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

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

    (If instead the cut were perpendicular to the length — into n shorter magnets — each would have moment M/n as well but for the opposite reason, shorter length with the same pole strength. Here, with cuts parallel to the length, the answer is M/n.)

    ✓Final answer

    (A) M/n.

12. 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 $\vec{B}$:

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

    with magnitude $\tau = mB\sin\theta$, where $\theta$ is the angle between $\vec{m}$ and $\vec{B}$. The torque is zero when $\vec{m}$ is aligned (or anti-aligned) with $\vec{B}$, and maximum when they are perpendicular. This is exactly analogous to the torque on an electric dipole, $\vec{\tau} = \vec{p}\times\vec{E}$. It is a cross product (giving a vector perpendicular to both m and B), not a dot product or a sum, so options (a), (c), (d) are dimensionally/conceptually wrong.

    ✓Final answer

    (b) tau = m x B.