$A$ long straight conductor is bent into the shape as shown. If it carries a current $i$ and the radius of the circular part is $R$,then find the magnetic field $B$ at the centre of the circular coil.

Two long parallel wires $X$ and $Y$,separated by a distance of $6 \text{ cm}$,carry currents of $5 \text{ A}$ and $4 \text{ A}$,respectively,in opposite directions as shown in the figure. The magnitude of the resultant magnetic field at point $P$,which is at a distance of $4 \text{ cm}$ from wire $Y$,is $x \times 10^{-5} \text{ T}$. The value of $x$ is . . . . . . .
Take the permeability of free space as $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.

What is the ratio of the magnetic field at point $O$ in the given figures?

An electron moves in a circular orbit with uniform speed $v$. It produces a magnetic field $B$ at the centre of the circle. The radius of the circle is (where $\mu_{0} =$ permeability of free space,$e =$ electronic charge):

Two long parallel wires separated by $0.1 \ m$ carry currents of $1 \ A$ and $2 \ A$, respectively, in opposite directions. $A$ third current-carrying wire parallel to both of them is placed in the same plane such that it feels no net magnetic force. It is placed at a distance of

Two long,straight wires carry equal currents of $10 \ A$ in opposite directions as shown in the figure. The separation between the wires is $5.0 \ cm$. The magnitude of the magnetic field at a point $P$ midway between the wires is . . . . . . $\mu T$. (Given: $\mu_0 = 4\pi \times 10^{-7} \ TmA^{-1}$)