Magnetic field intensity at the centre of a coil of $50$ turns,radius $0.5\, m$ and carrying a current of $2\, A$ 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

$A$ wire bent in the shape of a regular $n$-polygonal loop carries a steady current $I$. Let $l$ be the perpendicular distance of a given segment and $R$ be the distance of a vertex from the centre of the loop. The magnitude of the magnetic field at the centre of the loop is given by

The magnetic field due to a current in a straight wire segment of length $L$ at a point on its perpendicular bisector at a distance $r$ $(r >> L)$ 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 $(\mu_0 = \text{permeability of free space})$:

Two circular coils $X$ (smaller) and $Y$ (bigger), each having a single turn, carry equal currents in the same direction and subtend the same angle at point '$O$' along the axis of the coil. The distances of the centers of coil to point '$O$' are $d$ and $(d/2)$ for coil $Y$ and $X$ respectively. The radii of coils $Y$ and $X$ are $(2r)$ and $(r)$ respectively. The magnetic induction due to the bigger coil at point '$O$' is $B_y$ and that due to smaller coil $X$ at point '$O$' is $B_x$. $(d \gg r)$ The relation between $B_x$ and $B_y$ is