$A$ straight wire carrying current $I$ is bent into a semi-circular arc of radius $r$,as shown. The magnitude of the magnetic field at point $O$ due to the semi-circular arc is ($\mu_{0} =$ permeability of free space).

In an atom, electrons revolve around the nucleus along a path of radius $0.72 \text{ Å}$, making $9.4 \times 10^{18}$ revolutions per second. The equivalent current is [given, $e = 1.6 \times 10^{-19} \text{ C}$]. (in $\text{A}$)

Two straight infinitely long current-carrying wires are kept along the $z-$axis at the coordinates $(0, a, 0)$ and $(0, -a, 0)$ respectively, as shown in the figure. The current in each wire is equal and directed along the negative $z-$axis (into the plane of the paper). The variation of the magnetic field on the $x-$axis will be approximately:

What should be the current $i$ (in $A$) in a circular coil of radius $5\,cm$ to annul the horizontal component of Earth's magnetic field ${B_H} = 5 \times {10^{ - 5}}\,T$?

As shown in the figure,a long straight conductor with a semicircular arc of radius $R = \frac{\pi}{10} \, m$ is carrying a current $I = 3 \, A$. The magnitude of the magnetic field at the center $O$ of the arc is $........... \mu T$. (The permeability of the vacuum $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$)

As shown in the figure,a current of $2\,A$ flows in an equilateral triangle of side $4 \sqrt{3}\,cm$. The magnetic field at the centroid $O$ of the triangle is:
(Neglect the effect of the earth's magnetic field.)