$A$ long straight wire carries an electric current of $2\,A$. The magnetic induction at a perpendicular distance of $5\,m$ from the wire is

$A$ wire of resistance $R$ is bent in the form of a square of side $a$ as shown in the figure. Find the magnetic induction at the center of the square $O$ due to the current flowing through it.

Two mutually perpendicular insulated conducting wires carrying equal currents $I$ intersect at the origin. The resultant magnetic induction at point $P(2 \ m, 3 \ m)$ will be:

Two identical circular wires of radius $20\,cm$ and carrying current $\sqrt{2}\,A$ are placed in perpendicular planes as shown in the figure. The net magnetic field at the centre of the circular wires is $.............\times 10^{-8}\,T$. (Take $\pi=3.14$ )

The magnetic field at a point $P$ situated at a perpendicular distance $R$ from a long straight wire carrying a current of $12 \ A$ is $3 \times 10^{-5} \ Wb/m^2$. The value of $R$ in $mm$ is $\left[\mu_0 = 4\pi \times 10^{-7} \ Wb/Am\right]$

Two long parallel wires carrying currents $8 \,A$ and $15 \,A$ in opposite directions are placed at a distance of $7 \,cm$ from each other. $A$ point $P$ is equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of the magnetic field at point $P$ is $(\sqrt{2}=1.4)$ $(\mu_0=4 \pi \times 10^{-7} \,T \cdot m/A)$.