$A$ horizontal overhead power line carries a current of $90 \text{ A}$ in east to west direction. What is the magnitude and direction of the magnetic field due to the current $1.5 \text{ m}$ above the line?

Match List-$I$ with List-$II$. Choose the correct answer from the options given below:

$A$ 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 $(\mu_0 = \text{permeability of free space})$:

To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region,one can use the saddle coils illustrated in the figure. The loops are wrapped over a somewhat flattened tube. Assume the straight sections of wire are very long. The end view of the tube shows how the windings are applied. The overall current distribution is the superposition of two overlapping circular cylinders of uniformly distributed current,one toward you and one away from you. The current density $J$ is the same for each cylinder. The position of the axis of one cylinder is described by a position vector $\vec{a}$ relative to the other cylinder. The magnetic field inside the hollow tube is:

$A$ current of $8 \text{ A}$ each flows in opposite directions in two parallel conducting wires placed at a distance of $30 \text{ cm}$. The magnitude of the magnetic field at the midpoint between the two wires is . . . . . . $\mu \text{T}$. (Given: $\frac{\mu_0}{4\pi} = 10^{-7} \text{ N/A}^2$)

Two circular coils $P$ and $Q$ are made from similar wire,but the radius of $Q$ is twice that of $P$. What should be the value of the potential difference across them so that the magnetic induction at their centres is the same?