The dimensional formula of magnetic flux is

The correct option is [ML²T⁻²A⁻¹].

To find the dimensional formula of magnetic flux, we can use its defining physical relationship:
$\Phi =B·A$
where:
• $\Phi$ is the magnetic flux,
• $B$ is the magnetic field strength (magnetic induction), and
• $A$ is the area.

First, let us determine the dimensional formula for each component:

1. Area (A):
The dimensional formula for area is:
$\left[A\right]=\left[L^2\right]$

2. Magnetic Field (B):
We can find the dimensions of the magnetic field from the magnetic force equation acting on a moving charge:
$F=q·v·B$
Rearranging for $B$, we get:
$B=\frac{F}{q·v}$
Here:
• Force ($F$) has dimensions: $\left[F\right]=\left[ML{T}^{-2}\right]$
• Charge ($q$) is electric current multiplied by time: $\left[q\right]=\left[AT\right]$ (where $A$ represents Ampere)
• Velocity ($v$) has dimensions: $\left[v\right]=\left[L{T}^{-1}\right]$

Substituting these dimensions into the expression for $B$:
$\left[B\right]=\frac{\left[ML{T}^{-2}\right]}{\left[AT\right]·\left[L{T}^{-1}\right]}$
Simplifying the denominator:
$\left[AT\right]·\left[L{T}^{-1}\right]=\left[AL\right]$
Thus, the dimensions of $B$ are:
$\left[B\right]=\frac{\left[ML{T}^{-2}\right]}{\left[AL\right]}=\left[M{T}^{-2}{A}^{-1}\right]$

3. Magnetic Flux ($\Phi$):
Now, substitute the dimensions of $B$ and $A$ into the formula for magnetic flux:
$\left[\Phi \right]=\left[B\right]·\left[A\right]$
$\left[\Phi \right]=\left[M{T}^{-2}{A}^{-1}\right]·\left[L^2\right]$
$\left[\Phi \right]=\left[M{L}^2{T}^{-2}{A}^{-1}\right]$

Therefore, the dimensional formula of magnetic flux is [ML²T⁻²A⁻¹].