The dimension of magnetic induction can be written as

The correct answer is Option (1) → \( [M^{1}L^{0}T^{-2}A^{-1}] \)

Magnetic induction \( B \) is defined through the magnetic (Lorentz) force on a moving charge, \( F = qvB \) for velocity perpendicular to the field.

    \( \displaystyle B = \frac{F}{qv} \)

    \( [F] = [MLT^{-2}] \)
    \( [q] = [AT] \)
    \( [v] = [LT^{-1}] \)

    \( \displaystyle [B] = \frac{[MLT^{-2}]}{[AT][LT^{-1}]} = [M^{1}L^{0}T^{-2}A^{-1}] \)

The same result follows from the unit: \( 1\,\mathrm{T} = 1\,\mathrm{N\,A^{-1}\,m^{-1}} \), and \( \mathrm{N\,A^{-1}\,m^{-1}} = \mathrm{kg\,m\,s^{-2}\,A^{-1}\,m^{-1}} = \mathrm{kg\,s^{-2}\,A^{-1}} \).

NCERT (2026-27):
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