The strength of the magnetic field at a point $r$ near a long straight current-carrying wire is $B$. The field at a distance $\frac{r}{2}$ will be

The current passing through a conducting loop in the form of an equilateral triangle of side $4\sqrt{3} \text{ cm}$ is $2 \text{ A}$. The magnetic field at its centroid is $\alpha \times 10^{-5} \text{ T}$. The value of $\alpha$ is . . . . . . . (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

It is found that a non-zero current element is unable to produce any magnetic field at a particular point. Then the angle between the current element and the position vector of that point with respect to the current element is

$A$ 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 $.........\,\times 10^{-4}\,T$.

The magnetic field $d\overrightarrow{B}$ due to a small current element $d\overrightarrow{l}$ at a distance $\overrightarrow{r}$ from an element carrying current $i$ is given by:

Two parallel conducting wires of equal length are placed distance $d$ apart and carry currents $I_1$ and $I_2$ respectively in opposite directions. The resultant magnetic field at the midpoint of the distance between both the wires is