Two long straight parallel conductors carry currents $I_1$ and $I_2$ $(I_1 > I_2)$. When the direction of $I_1$ and $I_2$ is the same,the magnetic field intensity midway between the two conductors is $20 \mu T$. If the direction of $I_2$ is reversed,the field intensity becomes $50 \mu T$. The ratio $I_2 / I_1$ is:

$PQ$ and $RS$ are long parallel conductors separated by a certain distance. $M$ is the midpoint between them (see the figure). The net magnetic field at $M$ is $B$. Now,the current $2 \text{ A}$ is switched off. The field at $M$ now becomes

$A$ length $L$ of wire carries a steady current $I$. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is

$A$ long curved conductor carries a current $I$. $A$ small current element of length $dl$ on the wire induces a magnetic field at a point away from the current element. If the position vector between the current element and the point is $\vec{r}$, making an angle $\theta$ with the current element, then the induced magnetic field density $d\vec{B}$ at the point is $(\mu_0 = \text{permeability of free space})$:

The magnetic field at the centre of a current-carrying circular loop of radius $R$ is $16 \ \mu T$. The magnetic field at a distance $x = \sqrt{3}R$ on its axis from the centre is . . . . . . $\mu T$.

Two circular coils $1$ and $2$ are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$,so that the magnetic field at their centre is the same?