SI unit of magnetic flux is
(A) $ampere - meter$
(B) $ampere - {m^2}$
(C) $weber$
(D) $\dfrac{{weber}}{{{m^2}}}$

Hint:Here, we are asked to find the SI unit of magnetic flux. In order to proceed, you must know the definition of the magnetic flux. The magnetic flux is a term which is a combination or say an expression of magnetic field and area. Consider this expression, combine the SI units of magnetic field and area and then find the SI unit of magnetic flux.

Complete answer:
Consider a certain magnetic field whose magnetic strength is equal to $B$. The corresponding magnetic field vector field be ${\mathbf{B}}$. Now, imagine a surface of a certain area $A$, in that surface, consider an infinitesimal area $dA$. If you integrate the normal component of the magnetic field over the surface $dA$, you get the surface integral which is called as the magnetic flux through the surface of area $A$.

Basically, it is the number of magnetic field lines that passes through that surface area. Mathematically, the normal component of the magnetic field over the surface area $dA$ can be given as a dot product of $B$ and $dA$. We denote magnetic flux by the letter phi, so it will be given as $\Phi = \int {{\mathbf{B}}.d{\mathbf{A}}} $. The SI unit of magnetic field is Tesla denoted as $T$and the SI unit of area is square meter and is denoted as ${m^2}$.

If the product of these two is taken, the resulting quantity will have the SI unit as the product of the SI units of these two. Hence, the SI unit of magnetic flux is Tesla-square meter which is $T - {m^2}$. Now, Tesla- square meter is actually weber.Therefore, the SI unit of magnetic flux is $weber$.

Hence, option C is correct.

Note: Remember that the magnetic flux is defined as the number of magnetic field lines passing through an area and can be found out by taking the dot product of the magnetic field vector and the area vector. Also keep in mind that $weber$ is equivalent to $Tesla - {m^2}$. From there you can easily remember the formula of magnetic flux.