Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Physical Quantities | Dimensional Formula | ||
| (A) | Permeability of free space | (I) | \( [MT^{-2}A^{-1}] \) |
| (B) | Magnetic moment | (II) | \( [ML^{2}T^{-2}A^{-1}] \) |
| (C) | Magnetic flux | (III) | \( [MLT^{-2}A^{-2}] \) |
| (D) | Magnetic flux density | (IV) | \( [L^{2}A] \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) - (III), (B) - (IV), (C) - (II), (D) - (I)
Write each quantity through a relation whose dimensions are known, using force \( [MLT^{-2}] \), energy \( [ML^{2}T^{-2}] \) and current \( [A] \).
(A) Permeability of free space — this matches (III) \( [MLT^{-2}A^{-2}] \). From the force per unit length between two parallel currents, \( F/l = \mu_{0}I_{1}I_{2}/(2\pi d) \); the lengths \( l \) and \( d \) cancel in dimensions, leaving force divided by current squared.
\( [\mu_{0}] = \frac{[MLT^{-2}]}{[A^{2}]} = [MLT^{-2}A^{-2}] \)
(B) Magnetic moment — this matches (IV) \( [L^{2}A] \). For a current loop, magnetic moment = current × area, \( m = IA \).
(C) Magnetic flux — this matches (II) \( [ML^{2}T^{-2}A^{-1}] \). Flux is field × area: \( [MT^{-2}A^{-1}][L^{2}] \).
(D) Magnetic flux density — this matches (I) \( [MT^{-2}A^{-1}] \). From the force on a current-carrying wire, \( F = BIl \), so \( B = F/(Il) \).
\( [B] = \frac{[MLT^{-2}]}{[A][L]} = [MT^{-2}A^{-1}] \)
NCERT (2026-27):
Summary – …
