What is the ratio of the electric field to the magnetic field produced by a moving point charge if its speed is $4.5 \times 10^{5} \; m/s$?

$A$ very long wire carrying a current $I = 4 \sqrt{2} \, A$ is bent at a right angle. The magnitude of the magnetic field $(|B|)$ at a point $P$ lying on a line perpendicular to the bent wire at a distance $d = 20 \, cm$ from the point of the bending will be (Let $\mu_0 = 4 \pi \times 10^{-7} \, H/m$): (in $\, \mu T$)

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 $\pi=3.14$ )

Give similarity between Biot-Savart law and electrostatic law of Coulomb.

$AB$ and $CD$ are long straight conductors,separated by a distance $d$,each carrying a current $I$. The magnetic field at the midpoint of $BC$ is

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-