$Weber/m^2$ is equal to

The direction of magnetic lines of force produced by passing a direct current in a conductor is given by

Due to $10 \ A$ of current flowing in a circular coil of $10 \ cm$ radius,the magnetic field produced at its centre is $3.14 \times 10^{-3} \ Wb/m^2$. The number of turns in the coil will be:

Two concentric coplanar circular loops of radii $r_1$ and $r_2$ carry currents of respectively $i_1$ and $i_2$ in opposite directions (one clockwise and the other anticlockwise). The magnetic induction at the centre of the loops is half that due to $i_1$ alone at the centre. If $r_2 = 2r_1$,the value of $i_2/i_1$ is:

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the mid-point between them. The net magnetic field at $M$ is $B$. Now,the current $2\,A$ is switched off. The field at $M$ now becomes

The electric current in a circular coil of $2$ turns produces a magnetic induction $B_{1}$ at its centre. The coil is unwound and is rewound into a circular coil of $5$ turns and the same current produces a magnetic induction $B_{2}$ at its centre. The ratio of $\frac{B_{2}}{B_{1}}$ is: