$A$ current carrying circular coil of radius $R$ produces magnetic field $B_1$ at an axial point $P$ at a distance $x$ from its centre and $B_2$ at point $Q$ placed at its centre respectively. If $B_2 = 8B_1$, the value of $x$ is

For the magnetic field to be maximum due to a small element of a current-carrying conductor at a point,the angle between the element and the line joining the element to the given point must be.......$^o$

The magnetic induction at the centre $O$ of the current-carrying bent wire shown in the adjoining figure is:

Magnetic field at a distance $r$ from an infinitely long straight conductor carrying a steady current varies as

Current through $ABC$ and $A^{\prime} B^{\prime} C^{\prime}$ is $I$. What is the magnetic field at $P$? $BP = PB^{\prime} = r$. (Here $C^{\prime} B^{\prime} PBC$ are collinear).

Two long parallel wires carrying currents $8 \text{ A}$ and $15 \text{ A}$ in opposite directions are placed at a distance of $7 \text{ cm}$ from each other. $A$ point $P$ is equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnetic field at $P$ is $X \times 10^{-6} \text{ T}$. Find $X$. (Given: $\sqrt{2} = 1.4, \mu_0 = 4\pi \times 10^{-7} \text{ SI unit}$)