$A$ and $B$ are two concentric circular conductors with center $O$,carrying currents $i_1$ and $i_2$ as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the magnetic flux densities at $O$ due to $A$ and $B$ is $1:3$,then the value of $i_1/i_2$ is:

In the given diagram,two current-carrying circular loops of radius $R$ and $2R$ are arranged in the $YZ-$ plane and $XZ-$ plane respectively. The common center of both is at the origin $O$. What will be the angle of the resultant magnetic field from the $X-$ axis?

$PQRS$ is a square loop made of uniform conducting wire. If the current enters the loop at $P$ and leaves at $S$, then the magnetic field will be

Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre,the ratio of $i_1: i_2$ is

Two concentric coils each of radius equal to $2\pi \text{ cm}$ are placed at right angles to each other. If $3 \text{ A}$ and $4 \text{ A}$ are the currents flowing through the two coils respectively,the magnetic induction (in $\text{Wb m}^{-2}$) at the centre of the coils will be:

An infinitely long conductor $PQR$ is bent to form a right angle as shown. $A$ current $I$ flows through $PQR$. The magnetic field due to this current at the point $M$ is $H_1$. Now,another infinitely long straight conductor $QS$ is connected at $Q$ so that the current is $I/2$ in $QR$ as well as in $QS$,while the current in $PQ$ remains unchanged. The magnetic field at $M$ is now $H_2$. The ratio $H_1/H_2$ is given by