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).

$A$ current-carrying circular loop of radius '$R$' and a current-carrying long straight wire are placed in the same plane. The currents through the circular loop and the long straight wire are '$I_C$' and '$I_w$' respectively. The perpendicular distance between the centre of the circular loop and the wire is '$d$'. The magnetic field at the centre of the loop will be zero when the separation '$d$' is equal to:

$A$ straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ of the loop is

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:

What should be the current $i$ (in $A$) in a circular coil of radius $5\,cm$ to annul the horizontal component of Earth's magnetic field ${B_H} = 5 \times {10^{ - 5}}\,T$?

$A$ circular current-carrying coil has radius $R$. At what distance from the centre of the coil on the axis,the magnetic induction will become $\frac{1}{8}$ th of its value at the centre of the coil?