$A$ square loop of side $1\,m$ and resistance $1\,\Omega$ is placed in a magnetic field of $0.5\,T$. If the plane of the loop is perpendicular to the direction of the magnetic field,the magnetic flux through the loop is $\dots\dots$ weber.

$A$ circular disc of radius $0.2 \; m$ is placed in a uniform magnetic field of induction $\frac{1}{\pi} \; Wb/m^2$ in such a way that its axis makes an angle of $60^{\circ}$ with $\vec{B}$. The magnetic flux linked with the disc is..... $Wb$.

$A$ square of side $L$ metre lies in the $x-y$ plane in a region where the magnetic field is $\vec{B} = B_0(2 \hat{i} + 3 \hat{j} + 4 \hat{k})$,where $B_0$ is a constant. The magnitude of the magnetic flux passing through the square (in weber) is: (in $B_0 L^2$)

$A$ square of side $x \, m$ lies in the $x-y$ plane in a region where the magnetic field is given by $\vec B = B_0 (3\hat i + 4\hat j + 5\hat k ) \, T$,where $B_0$ is a constant. The magnitude of the magnetic flux passing through the square is:

Two circular loops of diameters $0.6 \,cm$ and $40 \,cm$ are kept coaxially with a separation of $15 \,cm$ between their centres. If a current $2 \,A$ flows through the smaller loop, then the flux linked with the bigger loop is (approximately)

$A$ uniform magnetic field of strength $B=2 \text{ mT}$ exists vertically downwards. These magnetic field lines pass through a closed surface as shown in the figure. The closed surface consists of a hemisphere $S_1$,a right circular cone $S_2$,and a circular surface $S_3$. The magnetic flux through $S_1$ and $S_2$ are respectively: