The net magnetic flux through any closed surface, kept in uniform magnetic field is:

A circular disc of radius \(0.2\) m is placed in a uniform magnetic field of induction \(\frac{1}{\pi} \left(\frac{\text{Wb}}{\text{m}^{2}}\right)\) in such a way that its axis makes an angle of \(60^{\circ}\) with \(\vec {B}.\) The magnetic flux linked to the disc will be:

If a current is passed through a circular loop of radius \(R\) then magnetic flux through a coplanar square loop of side \(l\) as shown in the figure \((l<<R)\) is:

The radius of a loop as shown in the figure is \(10~\text{cm}.\) If the magnetic field is uniform and has a value \(10^{-2}~ \text{T},\) then the flux through the loop will be:
 

A square of side \(L\) meters lies in the \(XY\text-\)plane in a region where the magnetic field is given by \(\vec{B}=B_{0}\left ( 2\hat{i} +3\hat{j}+4\hat{k}\right )\text{T}\) where \(B_{0}\) is constant. The magnitude of flux passing through the square will be:
1. \(2 B_{0} L^{2}~\text{Wb}\)
2. \(3 B_{0} L^{2}~\text{Wb}\)
3. \(4 B_{0} L^{2}~\text{Wb}\)
4. \(\sqrt{29} B_{0} L^{2}~\text{Wb}\)$$

What is the dimensional formula of magnetic flux?
1. \(\left[ M L^2 T^{-2}A^{-1}\right]\)
2. \(\left[ M L^1 T^{-1}A^{-2}\right]\)
3. \(\left[ M L^2 T^{-3}A^{-1}\right]\)
4. \(\left[ M L^{-2} T^{-2}A^{-2}\right]\)$$