$A$ charged particle $A$ with charge $q = 2 \, C$ has a velocity $v = 100 \, m/s$. When it passes through point $A$ with the velocity in the direction shown in the figure,the strength of the magnetic field at point $B$ due to this moving charge is ....... $\mu T$ $(r = 2 \, m)$.

Define the law for finding the direction of a magnetic field due to a circular current loop.

$A$ straight wire carrying a current of $14\,A$ is bent into a semicircular arc of radius $2.2\,cm$ as shown in the figure. The magnetic field produced by the current at the centre $(O)$ of the arc is $.........\,\times 10^{-4}\,T$.

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 $P$ to the wires are perpendicular to each other. The magnitude of magnetic field at $P$ is $............\times 10^{-6}\,T$. (Given : $\sqrt{2}=1.4$)

An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a current $I = 10 \,A$. The magnetic field on the $y$-axis at a distance of $0.5 \,m$ from the element of length $\Delta x = 1 \,cm$ is:

$A$ square frame of side $l$ carries a current $i$. The magnetic field at its centre is $B$. The same current is passed through a circular coil having the same perimeter as the square. The field at the centre of the circular coil is $B^{\prime}$. The ratio of $\frac{B}{B^{\prime}}$ is