The dimensions of magnetic flux are

The correct answer is Option (4) → \( [\mathrm{ML^{2}T^{-2}A^{-1}}] \)

Magnetic flux through an area \( A \) in a uniform field is \( \Phi_B=BA\cos\theta \). So its dimensions are those of the magnetic field times an area. We get the dimensions of \( B \) from the magnetic force \( F=qvB \) on a moving charge. The charge is written as current × time.

The dimensions of the quantities in \( F=qvB \) are:

    \( [F]=[\mathrm{MLT^{-2}}] \)

    \( [q]=[\mathrm{AT}] \)

    \( [v]=[\mathrm{LT^{-1}}] \)

The dimensions of the magnetic field are:

    \( \begin{aligned} [B] &= \left[\frac{F}{qv}\right] \\ &= \frac{[\mathrm{MLT^{-2}}]}{[\mathrm{AT}]\,[\mathrm{LT^{-1}}]} \\ &= [\mathrm{MT^{-2}A^{-1}}] \end{aligned} \)

Since \( \cos\theta \) has no dimensions, the dimensions of the flux are:

    \( \begin{aligned} [\Phi_B] &= [B]\,[\mathrm{L^{2}}] \\ &= [\mathrm{MT^{-2}A^{-1}}]\times[\mathrm{L^{2}}] \\ &= [\mathrm{ML^{2}T^{-2}A^{-1}}] \end{aligned} \)

NCERT (2026-27):
Magnetic Flux – “… Magnetic flux through a plane of area A placed in a uniform magnetic field B … can be written as

    \( \Phi_B=\mathbf{B}\cdot\mathbf{A}= \)\( BA\cos\theta \)\( \qquad (6.1) \)

where θ is angle between B and A.”

Magnetic Field, Lorentz Force – “… Dimensionally, we have [B] = [F/qv] …”