Two identical long current-carrying wires are bent into the shapes shown in the following figures. If the magnitudes of the magnetic fields at the centers $P$ and $Q$ of the semicircular arcs are $B_1$ and $B_2$ respectively,then the ratio $\frac{B_1}{B_2}$ is . . . . . . .

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \; cm$,carries a current of $0.40 \; A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?

If we double the radius of a coil keeping the current through it unchanged,then the magnetic field at any point at a large distance from the centre becomes approximately

$A$ wire in the form of a circular loop of one turn carrying a current produces a magnetic field $B$ at the centre. If the same wire is looped into a coil of two turns and carries the same current,the new value of magnetic induction at the centre is

Two concentric circular coils $X$ and $Y$ of radii $16\; cm$ and $10\; cm$ respectively,lie in the same vertical plane containing the north to south direction. Coil $X$ has $20$ turns and carries a current of $16\; A$. Coil $Y$ has $25$ turns and carries a current of $18\; A$. The sense of the current in $X$ is anticlockwise,and clockwise in $Y$,for an observer looking at the coils facing west. Give the magnitude and direction of the net magnetic field due to the coils at their centre.

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)