$A$ long straight wire of radius $a$ carries a steady current $i$. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at $a/2$ and $2a$ is:

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

Two concentric circular coils $A$ and $B$ having radii $20 \ cm$ and $10 \ cm$ respectively lie in the same plane. The current in coil $A$ is $0.5 \ A$ in anticlockwise direction. The current in coil $B$,so that net magnetic field at the common centre is zero,is

As shown in the figure,two infinitely long straight parallel wires $P$ and $Q$ carrying equal currents in opposite directions are arranged parallel to the $Y$-axis. If the magnetic field due to wire $P$ at the origin '$O$' of the coordinate system is $B$,then match the resultant magnetic fields at various points given in Column $A$ with the points given in Column $B$.

Column $A$	Column $B$
$A) \frac{B}{4}$	$i) (0, 0)$
$B) \frac{B}{2}$	$ii) (a, 0)$
$C) \frac{2B}{3}$	$iii) (2a, 0)$
$D) 2B$	$iv) (3a, 0)$

$A$ wire of length $L = 1 \ m$ carries a constant current $I$. The wire is bent to form a single circular loop of radius $R$. The magnetic field at the centre of this loop is $B$. If the same wire is now bent to form a circular loop with $n = 4$ turns,what will be the magnetic field $B'$ at the centre of this new loop?

Which rule can be used to determine the direction of the magnetic field due to a current-carrying loop?