Magnetic Dipole & Bar Magnet — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Magnetic Dipole & Bar Magnet MCQs with step-by-step solutions (21 questions). Part of Magnetic Materials. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Magnetic Dipole & Bar Magnet · easy · theory
The magnetic dipole moment of a bar magnet of pole strength $m$ and magnetic length $2l$ is:
A. $2ml^2$
B. $\dfrac{2l}{m}$
C. $m \times 2l$ ✓ Correct
D. $\dfrac{m}{2l}$
Solution: The moment is directed from the south pole to the north pole inside the magnet.
Q2 — Magnetic Dipole & Bar Magnet · easy · theory
The SI unit of magnetic dipole moment is:
A. $\text{A}/\text{m}$
B. $\text{A}\cdot\text{m}$
C. $\text{A}\cdot\text{m}^2$ ✓ Correct
D. $\text{Wb}\cdot\text{m}$
Solution: Equivalently it is joule per tesla, since $U = -MB$.
Q3 — Magnetic Dipole & Bar Magnet · easy · theory
Isolated magnetic poles (monopoles) do not exist. The experimental evidence is that:
A. A freely suspended magnet points north
B. Like poles repel each other
C. Breaking a magnet always produces two complete magnets, each with both poles ✓ Correct
D. A magnet loses its magnetism when heated
Solution: However finely a magnet is divided, each fragment is a complete dipole — magnetic field lines are always closed loops.
Q4 — Magnetic Dipole & Bar Magnet · medium · theory
The magnetic field at an axial point of a short bar magnet at distance $r$ is:
A. $\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$
B. $\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$ ✓ Correct
C. $\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^2}$
D. $\dfrac{\mu_0}{4\pi}\dfrac{M}{r^2}$
Solution: The axial (end-on) field falls off as the inverse cube of the distance, twice the equatorial value.
Q5 — Magnetic Dipole & Bar Magnet · medium · theory
The magnetic field at an equatorial point of a short bar magnet at distance $r$ is:
A. Zero
B. $\dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3}$
C. $\dfrac{\mu_0}{4\pi}\dfrac{M}{r^2}$
D. $\dfrac{\mu_0}{4\pi}\dfrac{M}{r^3}$ ✓ Correct
Solution: The equatorial (broadside-on) field is antiparallel to the moment and half the axial value at the same distance.
Q6 — Magnetic Dipole & Bar Magnet · easy · theory
The torque on a bar magnet of moment $M$ placed at angle $\theta$ in a uniform field $B$ is:
A. $\dfrac{MB}{\sin\theta}$
B. $MB\cos\theta$
C. $MB\tan\theta$
D. $MB\sin\theta$ ✓ Correct
Solution: It is greatest when the magnet lies across the field and vanishes when it is aligned with it.
Q7 — Magnetic Dipole & Bar Magnet · medium · theory
The potential energy of a bar magnet of moment $M$ in a uniform field $B$ at angle $\theta$ is:
A. $MB\tan\theta$
B. $-MB\cos\theta$ ✓ Correct
C. $+MB\cos\theta$
D. $-MB\sin\theta$
Solution: $U = -\vec{M}\cdot\vec{B}$, least when aligned and greatest when antiparallel.
Q8 — Magnetic Dipole & Bar Magnet · hard · theory
The magnetic length of a bar magnet is approximately:
A. Equal to its geometric length
B. Twice its geometric length
C. Five-sixths of its geometric length ✓ Correct
D. Half its geometric length
Solution: The effective poles lie a little inside the ends of the magnet, so the magnetic length is about $0.83$ times the geometric length.
Q9 — Magnetic Dipole & Bar Magnet · medium · numerical
A bar magnet of magnetic moment $M$ is cut along its length into two identical halves. The magnetic moment of each half is:
A. $\dfrac{M}{2}$ ✓ Correct
B. $2M$
C. $\dfrac{M}{4}$
D. $M$
Solution: Cutting lengthwise halves the pole strength while leaving the magnetic length unchanged, so $M' = \dfrac{m}{2} \times 2l = \dfrac{M}{2}$.
Q10 — Magnetic Dipole & Bar Magnet · hard · numerical
A bar magnet of moment $M$ is cut transversely into $n$ equal pieces. The moment of each piece is:
A. $\dfrac{M}{n}$ ✓ Correct
B. $nM$
C. $M$
D. $\dfrac{M}{n^2}$
Solution: A transverse cut keeps the pole strength but divides the length by $n$, so each piece has moment $\dfrac{M}{n}$.
Q11 — Magnetic Dipole & Bar Magnet · medium · numerical
A bar magnet has pole strength $10\text{ A}\cdot\text{m}$ and magnetic length $0.1\text{ m}$. Its magnetic moment is:
A. $0.1\text{ A}\cdot\text{m}^2$
B. $100\text{ A}\cdot\text{m}^2$
C. $1\text{ A}\cdot\text{m}^2$ ✓ Correct
D. $10\text{ A}\cdot\text{m}^2$
Solution: $M = m \times 2l = 10 \times 0.1 = 1\text{ A}\cdot\text{m}^2$.
Q12 — Magnetic Dipole & Bar Magnet · medium · numerical
A magnet of moment $2\text{ A}\cdot\text{m}^2$ makes $30^\circ$ with a field of $0.5\text{ T}$. The torque on it is:
A. $0.5\text{ N}\cdot\text{m}$ ✓ Correct
B. $0.87\text{ N}\cdot\text{m}$
C. $0.25\text{ N}\cdot\text{m}$
D. $1\text{ N}\cdot\text{m}$
Solution: $\tau = MB\sin 30^\circ = 2 \times 0.5 \times 0.5 = 0.5\text{ N}\cdot\text{m}$.
Q13 — Magnetic Dipole & Bar Magnet · easy · numerical
A magnet of moment $0.5\text{ A}\cdot\text{m}^2$ is placed in a field of $0.2\text{ T}$. The maximum torque on it is:
A. $0.1\text{ N}\cdot\text{m}$ ✓ Correct
B. $2.5\text{ N}\cdot\text{m}$
C. $0.25\text{ N}\cdot\text{m}$
D. $0.7\text{ N}\cdot\text{m}$
Solution: $\tau_{max} = MB = 0.5 \times 0.2 = 0.1\text{ N}\cdot\text{m}$.
Q14 — Magnetic Dipole & Bar Magnet · hard · numerical
A short bar magnet of moment $1\text{ A}\cdot\text{m}^2$ produces an axial field at $0.2\text{ m}$ of:
A. $2.5 \times 10^{-6}\text{ T}$
B. $2.5 \times 10^{-5}\text{ T}$ ✓ Correct
C. $1.25 \times 10^{-5}\text{ T}$
D. $5 \times 10^{-5}\text{ T}$
Solution: $B = \dfrac{\mu_0}{4\pi}\dfrac{2M}{r^3} = \dfrac{10^{-7} \times 2 \times 1}{8 \times 10^{-3}} = 2.5 \times 10^{-5}\text{ T}$.
Q15 — Magnetic Dipole & Bar Magnet · hard · numerical
For the same magnet ($M = 1\text{ A}\cdot\text{m}^2$), the equatorial field at $0.2\text{ m}$ is:
A. $5 \times 10^{-5}\text{ T}$
B. $2.5 \times 10^{-5}\text{ T}$
C. $1.25 \times 10^{-5}\text{ T}$ ✓ Correct
D. $6.25 \times 10^{-6}\text{ T}$
Solution: $B = \dfrac{10^{-7} \times 1}{8 \times 10^{-3}} = 1.25 \times 10^{-5}\text{ T}$, exactly half the axial value.
Q16 — Magnetic Dipole & Bar Magnet · medium · numerical
At the same distance from a short bar magnet, the ratio of the axial field to the equatorial field is:
A. $1 : 1$
B. $4 : 1$
C. $2 : 1$ ✓ Correct
D. $1 : 2$
Solution: Comparing $\dfrac{2M}{r^3}$ with $\dfrac{M}{r^3}$ gives the ratio $2 : 1$.
Q17 — Magnetic Dipole & Bar Magnet · medium · numerical
If the distance from a short bar magnet is doubled, the axial field becomes:
A. One-sixteenth
B. One-eighth ✓ Correct
C. One-fourth
D. One-half
Solution: $B \propto \dfrac{1}{r^3}$, so doubling $r$ divides the field by $8$.
Q18 — Magnetic Dipole & Bar Magnet · medium · numerical
A magnet of moment $2\text{ A}\cdot\text{m}^2$ lies aligned with a field of $0.1\text{ T}$. Its potential energy is:
A. $+0.2\text{ J}$
B. Zero
C. $-0.1\text{ J}$
D. $-0.2\text{ J}$ ✓ Correct
Solution: $U = -MB\cos 0^\circ = -2 \times 0.1 = -0.2\text{ J}$, the minimum energy position.
Q19 — Magnetic Dipole & Bar Magnet · hard · numerical
The work done in turning the magnet of the previous type ($M = 2$, $B = 0.1\text{ T}$) from alignment to the antiparallel position is:
A. $0.2\text{ J}$
B. $0.1\text{ J}$
C. Zero
D. $0.4\text{ J}$ ✓ Correct
Solution: $W = U_{180^\circ} - U_{0^\circ} = (+MB) - (-MB) = 2MB = 0.4\text{ J}$.
Q20 — Magnetic Dipole & Bar Magnet · hard · numerical
Two identical bar magnets each of moment $M$ are placed at right angles to each other. The resultant magnetic moment is:
A. $M$
B. Zero
C. $2M$
D. $M\sqrt{2}$ ✓ Correct
Solution: Magnetic moments add as vectors, so for perpendicular moments the resultant is $\sqrt{M^2 + M^2} = M\sqrt{2}$.
Q21 — Magnetic Dipole & Bar Magnet · easy · numerical
A bar magnet of magnetic length $0.1\text{ m}$ has pole strength $20\text{ A}\cdot\text{m}$. Its magnetic moment is:
A. $2\text{ A}\cdot\text{m}^2$ ✓ Correct
B. $0.2\text{ A}\cdot\text{m}^2$
C. $20\text{ A}\cdot\text{m}^2$
D. $200\text{ A}\cdot\text{m}^2$
Solution: $M = m \times 2l = 20 \times 0.1 = 2\text{ A}\cdot\text{m}^2$.