The magnetic induction at the centre $O$ of the current-carrying bent wire shown in the adjoining figure is

$A$ 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

Two concentric coils $X$ and $Y$ of radii $16 \, cm$ and $10 \, cm$ lie in the same vertical plane containing $N-S$ direction. Coil $X$ has $20$ turns and carries $16 \, A$. Coil $Y$ has $25$ turns and carries $18 \, A$. Coil $X$ has current in anticlockwise direction and coil $Y$ has current in clockwise direction for an observer looking at the coils facing the west. The magnitude of the net magnetic field at their common centre is:

In a hydrogen atom,an electron moves in a circular orbit of radius $5.2 \times 10^{-11} \, m$ and produces a magnetic induction of $12.56 \, T$ at its nucleus. The current produced by the motion of the electron will be (Given $\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m$)

An element $\overrightarrow{\Delta \ell} = \Delta x \hat{i}$ is placed at the origin and carries a current of $10 \ A$. Find the magnitude of the magnetic field on the $Y$-axis at a distance of $0.5 \ m$,given $\Delta x = 1 \ cm$. (Use $\frac{\mu_0}{4 \pi} = 10^{-7} \ T \cdot m/A$)

Consider a long straight wire of a circular cross-section (radius $a$) carrying a steady current $I$. The current is uniformly distributed across this cross-section. The distances from the centre of the wire's cross-section at which the magnetic field [inside the wire,outside the wire] is half of the maximum possible magnetic field,anywhere due to the wire,will be