Two infinite wires bent in the form of an $L$ shape carry current $I$ as shown. What is the magnetic field at $O$?

According to the Biot-Savart law,what is the magnetic field at a point on the axis of a current-carrying wire?

Two identical wires, carrying equal currents are bent into circular coils $A$ and $B$ with $2$ and $3$ turns respectively. The ratio of the magnetic fields at the centres of the coils $A$ and $B$ is

An infinitely long straight conductor is bent into the shape as shown in the figure. It carries a current of $i$ $ampere$ and the radius of the circular loop is $r$ $metre$. Then the magnetic induction at its centre $O$ will be:

Two identical long parallel wires carry currents $I_1$ and $I_2$ such that $I_1 > I_2$. When the currents are in the same direction,the magnetic field at a point midway between the wires is $6 \times 10^{-6} \ T$. If the direction of $I_2$ is reversed,the field becomes $3 \times 10^{-5} \ T$. The ratio $\left(\frac{I_1}{I_2}\right)$ is-

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 . . . . . . .