A short bar magnet has magnetic moment \(m\). Distances are measured from its centre. Take \(B_0=\dfrac{\mu_0}{4\pi}\dfrac{m}{r^{3}}\).
Match List-I with List-II.
| List-I (Point) | List-II (Magnitude of the field) |
|---|---|
| (A) On the axis, at distance \(r\) | (I) \(B_0/8\) |
| (B) On the equatorial line, at distance \(r\) | (II) \(2B_0\) |
| (C) On the axis, at distance \(2r\) | (III) \(B_0/4\) |
| (D) On the equatorial line, at distance \(2r\) | (IV) \(B_0\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(II), (B)-(IV), (C)-(III), (D)-(I)
For a short bar magnet, the field on the axis is twice the field on the equatorial line at the same distance. Both fields fall as \( 1/r^{3} \), so doubling the distance makes each field one-eighth.
(A) On the axis, at distance \( r \) — This matches (II) \( 2B_0 \). The axial field is \( B_A=\frac{\mu_0}{4\pi}\frac{2m}{r^{3}} \), which is \( 2B_0 \).
(B) On the equatorial line, at distance \( r \) — This matches (IV) \( B_0 \). In magnitude, the equatorial field is \( B_E=\frac{\mu_0}{4\pi}\frac{m}{r^{3}} \), which is \( B_0 \).
(C) On the axis, at distance \( 2r \) — This matches (III) \( B_0/4 \). At \( 2r \), the term \( r^{3} \) becomes \( 8r^{3} \). So the axial field is \( 2B_0/8 \), that is \( B_0/4 \).
(D) On the equatorial line, at distance \( 2r \) — This matches (I) \( B_0/8 \). At \( 2r \), the equatorial field becomes one-eighth of \( B_0 \), that is \( B_0/8 \).
NCERT (2026-27):
The electrostatic analog – “In particular, we can write down the equatorial field (BE) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet:”
\( \displaystyle B_E=-\frac{\mu_0 m}{4\pi r^{3}} \)\( \qquad (5.4) \)
The electrostatic analog – “Likewise, the axial field (BA) of a bar magnet for r >> l is:”
\( \displaystyle B_A=\frac{\mu_0}{4\pi}\frac{2m}{r^{3}} \)\( \qquad (5.5) \)