How can we determine the direction of the magnetic field using the Biot-Savart law?

The dimensions of the ratio of magnetic flux $(\phi)$ and permeability $(\mu)$ are

An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a large current $I = 10 \; A$ (Figure). What is the magnetic field on the $y$-axis at a distance of $0.5 \; m$? Given $\Delta x = 1 \; cm$.

$A$ particle is moving with velocity $\overrightarrow{v} = \hat{i} + 3\hat{j}$ and it produces an electric field at a point given by $\overrightarrow{E} = 2\hat{k}$. It will produce a magnetic field at that point equal to (all quantities are in $SI$ units):

Two parallel wires carry currents $i_1$ and $i_2$ $(i_1 > i_2)$. When the currents are in the same direction,the magnetic field at a point midway between the wires is $10 \, \mu T$. When the direction of $i_2$ is reversed,the magnetic field at that point becomes $30 \, \mu T$. What is the ratio $\frac{i_1}{i_2}$?

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is