Let $B_1$ be the magnitude of the magnetic field at the center of a circular coil of radius $R$ carrying current $I$. Let $B_2$ be the magnitude of the magnetic field at an axial distance $x$ from the center. For $x : R = 3 : 4$,the ratio $\frac{B_2}{B_1}$ is:

Consider a long straight wire of a circular cross-section (radius $a$) carrying a steady current $I$. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire,outside the wire] is half of the maximum possible magnetic field,anywhere due to the wire,will be

$A$ 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

$A$ wire of length $l$ is bent into a circular loop of radius $R$ and carries a current $I$. The magnetic field at the centre of the loop is $B$. The same wire is now bent into a double loop of equal radii. If both loops carry the same current $I$ and it is in the same direction,the magnetic field at the centre of the double loop will be

$PQ$ 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

Two identical long conducting wires $AOB$ and $COD$ are placed at a right angle to each other,with one above the other such that $O$ is their common point. The wires carry currents $I_1$ and $I_2$,respectively. Point $P$ lies at a distance $d$ from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at point $P$ will be: