Match List-I with List-II.
| List-I | List-II |
| (A) Magnetic field | (I) \( \mathrm{NA^{-2}} \) |
| (B) Magnetic flux | (II) \( \mathrm{NA^{-1}m^{-1}} \) |
| (C) Magnetic permeability | (III) \( \mathrm{Am^{2}} \) |
| (D) Magnetic moment | (IV) \( \mathrm{NmA^{-1}} \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (II), (B) - (IV), (C) - (I), (D) - (III)
Each unit follows from a defining relation of the quantity.
(A) Magnetic field — this matches (II) \( \mathrm{NA^{-1}m^{-1}} \). From the force on a current-carrying wire, \( F = BIl \), so \( B = F/(Il) \).
\( \displaystyle [B] = \frac{\mathrm{N}}{\mathrm{A\,m}} = \mathrm{N\,A^{-1}\,m^{-1}} \)
(B) Magnetic flux — this matches (IV) \( \mathrm{NmA^{-1}} \). Flux is field times area, \( \Phi = BA \).
\( \displaystyle [\Phi] = \mathrm{N\,A^{-1}\,m^{-1}} \times \mathrm{m^{2}} = \mathrm{N\,m\,A^{-1}} \)
(C) Magnetic permeability — this matches (I) \( \mathrm{NA^{-2}} \). From the force between two parallel currents per unit length, \( F/l = \mu_0 I_1 I_2/(2\pi d) \).
\( \displaystyle [\mu_0] = \frac{\mathrm{N\,m^{-1}} \times \mathrm{m}}{\mathrm{A^{2}}} = \mathrm{N\,A^{-2}} \)
(D) Magnetic moment — this matches (III) \( \mathrm{Am^{2}} \). The moment of a current loop is current times area, \( m = IA \).
Note: The same question was also asked in CUET (UG) 2026 on 14-05-2026 (Slot 2).