$A$ current of $1\,A$ is flowing through the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$. The magnetic field at the centre of the triangle will be

$A$ very long conducting wire is bent in a semicircular shape from $A$ to $B$ as shown in the figure. The magnetic field at point $P$ for the steady current configuration is given by:

Two infinitely long straight wires lie in the $xy$-plane along the lines $x=+R$ and $x=-R$. The wire located at $x=+R$ carries a constant current $I_1$ and the wire located at $x=-R$ carries a constant current $I_2$. A circular loop of radius $R$ is suspended with its centre at $(0,0, \sqrt{3} R)$ and in a plane parallel to the $xy$-plane. This loop carries a constant current $I$ in the clockwise direction as seen from above the loop. The current in the wire is taken to be positive if it is in the $+\hat{j}$ direction. Which of the following statements regarding the magnetic field $\vec{B}$ is (are) true?
$(A)$ If $I_1=I_2$, then $\vec{B}$ cannot be equal to zero at the origin $(0,0,0)$.
$(B)$ If $I_1 > 0$ and $I_2 < 0$, then $\vec{B}$ can be equal to zero at the origin $(0,0,0)$.
$(C)$ If $I_1 < 0$ and $I_2 > 0$, then $\vec{B}$ can be equal to zero at the origin $(0,0,0)$.
$(D)$ If $I_1=I_2$, then the $z$-component of the magnetic field at the centre of the loop is $\left(-\frac{\mu_0 I}{2 R}\right)$.

In the loop shown,the magnetic induction at the point $O$ is

Three rings,each having equal radius $R$,are placed mutually perpendicular to each other,and each has its centre at the origin of the coordinate system. If current $I$ is flowing through each ring,then the magnitude of the magnetic field at the common centre 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: