$A$ current of $i$ ampere is flowing in an equilateral triangle of side $a$. The magnetic induction at the centroid will be

The magnetic field on the axis of a circular loop of radius $R = 100\,cm$ carrying current $I = \sqrt{2}\,A$,at a point $x = 1\,m$ away from the centre of the loop is given by:

Two concentric circular loops,one of radius $R$ and the other of radius $2R$,lie in the $xy$-plane with the origin as their common center,as shown in the figure. The smaller loop carries current $I_1$ in the anti-clockwise direction and the larger loop carries current $I_2$ in the clockwise direction,with $I_2 > 2I_1$. $\vec{B}(x, y)$ denotes the magnetic field at a point $(x, y)$ in the $xy$-plane. Which of the following statement$(s)$ is(are) correct?
$(A)$ $\vec{B}(x, y)$ is perpendicular to the $xy$-plane at any point in the plane.
$(B)$ $|\vec{B}(x, y)|$ depends on $x$ and $y$ only through the radial distance $r = \sqrt{x^2 + y^2}$.
$(C)$ $|\vec{B}(x, y)|$ is non-zero at all points for $r$.
$(D)$ $\vec{B}(x, y)$ points normally outward from the $xy$-plane for all the points between the two loops.

$A$ current $I$ flows around a closed path in the horizontal plane as shown in the figure. The path consists of eight arcs with alternating radii $r$ and $2r$. Each segment of arc subtends an equal angle at the common centre $P$. The magnetic field produced by the current path at point $P$ is

$A$ long current-carrying wire produces a magnetic field of $1 \ T$ at a distance of $r$. What will be the magnetic field at distances of $(a)$ $\frac{r}{2}$,$(b)$ $2r$,and $(c)$ $3r$?

Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is