Units & Dimensions

JEE Main level, single correct, medium

25 of 149

Question 25 of 149

Match the LIST-I with LIST-II
\(\begin{array}{||c|l||c|l||} \hline & \text{List-I} & & \text{List-II} \\ \hline \text{A.} & \text{Magnetic induction} & \text{I.} & \mathrm{M\,L\,T^{-2}\,A^{-2}} \\ \hline \text{B.} & \text{Magnetic flux} & \text{II.} & \mathrm{M\,L^{2}\,T^{-2}\,A^{-2}} \\ \hline \text{C.} & \text{Magnetic permeability} & \text{III.} & \mathrm{M\,L^{0}\,T^{-2}\,A^{-1}} \\ \hline \text{D.} & \text{Self inductance} & \text{IV.} & \mathrm{M\,L^{2}\,T^{-2}\,A^{-1}} \\ \hline \end{array}\)
Choose the correct answer from the options given below:

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Answer: (B) A-III, B-IV, C-I, D-II

Match physical quantities with dimensional formulas using fundamental principles.
Magnetic Induction (A): From F=qvBF = qvB, [B]=MLT−2AT⋅LT−1=MT−2A−1=[B] = \frac{MLT^{-2}}{AT \cdot LT^{-1}} = MT^{-2}A^{-1} = III
Magnetic Flux (B): Φ=BA\Phi = BA, [Φ]=MT−2A−1⋅L2=ML2T−2A−1=[\Phi] = MT^{-2}A^{-1} \cdot L^2 = ML^2T^{-2}A^{-1} = IV
Magnetic Permeability (C): From B=μHB = \mu H with [H]=ALT−1[H] = ALT^{-1}, [μ]=MT−2A−1AL−1T−1=MLT−2A−2=[\mu] = \frac{MT^{-2}A^{-1}}{AL^{-1}T^{-1}} = MLT^{-2}A^{-2} = I
Self Inductance (D): From ε=LdIdt\varepsilon = L\frac{dI}{dt}, [L]=ML2T−3A−1AT−1=ML2T−2A−2=[L] = \frac{ML^2T^{-3}A^{-1}}{AT^{-1}} = ML^2T^{-2}A^{-2} = II
Correct matching: A-III, B-IV, C-I, D-II

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