$A$ portion of a conductive wire is bent in the form of a semicircle of radius $r$ as shown in the figure. At the centre $O$ of the semicircle,the magnetic induction will be:

$A$ long conducting wire carrying a current $I$ is bent at $120^{\circ}$ (see figure). The magnetic field $B$ at a point $P$ on the angle bisector of the bend at a distance $d$ from the bend is ($\mu_{0}$ is the permeability of free space):

$\frac{V \cdot s}{A \cdot m}$ is the unit of which physical quantity?

$A$ hairpin-like shape as shown in the figure is made by bending a long current-carrying wire. What is the magnitude of the magnetic field at point $P$,which lies at the center of the semicircle?

$A$ long wire lies along the $X$-axis and carries a current of $40 \, A$ in the positive $x$-direction. $A$ second long wire is perpendicular to the $xy$-plane, passes through the point $(3.0 \, m) \hat{j}$, and carries a current along the positive $z$-direction. If the magnitude of the resultant magnetic field at the point $(2.0 \, m) \hat{j}$ is $R=5 \times 10^{-6} \, T$, then the current in the second wire is (Permeability of free space, $\mu_0=4 \pi \times 10^{-7} \, T \cdot m/A$) (in $A$)

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