$ABCDE$ is a very long wire carrying current $i_1$ as shown. The portion $BCD$ is a semicircle with centre at $O$. Another very long wire passes through point $O$, perpendicular to the plane of $ABCDE$ and it carries current $i_2$ outside the plane of paper. The two wires,

Three distinct current carrying wires intersect a finite rectangular plane $ABCD$. The current in the left wire and the loop is $I_1$. The direction of current in left-most wire and right-most loop is downwards as shown in the figure. The current $I_2$ through the middle wire is adjusted so that the path integral of the total magnetic field along the perimeter of the rectangle is zero, that is, ${\oint }_{\mathrm{ABCDA}}\overrightarrow{\mathrm{B}}·\overrightarrow{\mathrm{dl}}=0.$ Then the current $I_2$ is

The figure shows a long coaxial cable in which a current $I$ flows down through the inner cylinder of radius $a$ and the same current flows back up through the outer cylinder of radius $2a$. The cylinders are insulated from each other and the current is uniformly distributed over the area of the cross-section in each cylinder. The strength of the magnetic field at a distance $\raisebox{1ex}{$3a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ from the axis of the cable is

A small coil $C$ with $N=200$ turns is mounted on one end of a balance beam and introduced between the poles of an elctromagnet as shown in the figure. The cross sectional area of coil is $A=1.0 {\mathrm{cm}}^2$, length of arm $OA$ of the balance beam is $l=30 \mathrm{cm}$. When there is no current in the coil the balance is in equilibrium. On passing a current $I=22 \mathrm{mA}$ through the coil the equilibrium is restored by putting the additional counter weight of mass $∆m = 60 \mathrm{mg}$ on the balance pan. Find the magnetic induction at the spot where coil is located.