The magnetic field near a current-carrying conductor is given by

$A$ square frame of side $1 \ m$ carrying a current $I$ produces a magnetic field $B$ at its centre. The same current is passed through a circular coil having the same perimeter as the square. The magnetic field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is

The wire loop $PQRSP$ formed by joining two semicircular wires of radii $R_1$ and $R_2$ carries a current $I$ as shown. The magnitude of the magnetic field at the centre '$O$' is

$A$ cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of center $O$ with angle $AOC = 60^o$. If $B_1$ and $B_2$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC$ and $ADC$ respectively,the ratio $B_1/B_2$ is

The magnetic field at a distance $r$ from a long straight wire carrying current $i$ is $0.4 \ T$. The magnetic field at a distance $2r$ is: (in $T$)

Two long parallel wires are separated by a distance of $2.50 \ cm$. The force per unit length that each wire exerts on the other is $4 \times 10^{-5} \ N \ m^{-1}$, and the wires repel each other. The current in one wire is $0.5 \ A$. What is the current in the second wire (in $A$)? (Take $\mu_0 = 4 \pi \times 10^{-7} \ S.I. \ \text{unit}$)