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Home/Practice questions/Physics
PhysicsElectromagnetic Induction

Match List-I with List-IIList-I (Physical quantity)List-II (Dimension…

Match List-I with List-II

List-I (Physical quantity)List-II (Dimensions)
(A) Mutual inductance(I) [ML2T-3A-1]
(B) Magnetic flux(II) [M0T0L2A]
(C) EMF(III) ML2T-2A-1
(D) Magnetic moment(IV) ML2T-2A-2

Choose the correct answer from the options given below:

Answer & explanation+

Correct answer: option 1

The correct answer is Option (1) → (A) - (IV), (B) - (III), (C) - (I), (D) - (II)

Build each dimension from a defining relation, starting from force [MLT−2], energy [ML2T−2] and charge [AT].

(A) Mutual inductance — this matches (IV) ML2T-2A-2. Mutual inductance is the flux linked per unit current, \( M = \phi/I \), so it has the dimensions of flux divided by current.

    \( \displaystyle [M] = \frac{[\mathrm{ML^{2}T^{-2}A^{-1}}]}{[\mathrm{A}]} = [\mathrm{ML^{2}T^{-2}A^{-2}}] \)

(B) Magnetic flux — this matches (III) ML2T-2A-1. Flux is \( \phi = BA \), and B follows from the force on a current, \( F = BIl \).

    \( \displaystyle [B] = \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{A}][\mathrm{L}]} = [\mathrm{MT^{-2}A^{-1}}] \)

    \( \displaystyle [\phi] = [\mathrm{MT^{-2}A^{-1}}][\mathrm{L^{2}}] = [\mathrm{ML^{2}T^{-2}A^{-1}}] \)

(C) EMF — this matches (I) [ML2T-3A-1]. EMF is work done per unit charge.

    \( \displaystyle [\varepsilon] = \frac{[\mathrm{ML^{2}T^{-2}}]}{[\mathrm{AT}]} = [\mathrm{ML^{2}T^{-3}A^{-1}}] \)

(D) Magnetic moment — this matches (II) [M0T0L2A]. A current loop has magnetic moment \( m = IA \), current times area, so its dimensions are [L2A].

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