Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| (Physical quantity) | (Unit) | ||
| A. | Magnetic field | I. | NA−2 |
| B. | Magnetic flux | II. | NA−1m−1 |
| C. | Magnetic permeability | III. | Am2 |
| D. | Magnetic moment | IV. | 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, written in terms of newton, ampere and metre.
(A) Magnetic field — this matches (II) NA−1m−1. From the force on a current-carrying wire, F = IlB, so B = F/(Il) and 1 T = 1 N A−1 m−1.
(B) Magnetic flux — this matches (IV) NmA−1. Flux is field × area: N A−1 m−1 × m2 = N m A−1, the weber.
(C) Magnetic permeability — this matches (I) NA−2. From the force per unit length between parallel currents, \( F/l = \mu_0I_1I_2/(2\pi d) \), so \( \mu_0 \) is measured in N A−2 (the same as T m A−1).
(D) Magnetic moment — this matches (III) Am2. For a current loop the moment is current × area, m = IA.
Note: The same question was also asked in CUET (UG) 2022 on 26-08-2022 (Slot 2).