$A$ circular loop of radius $r$ is carrying current $I \ A$. The ratio of the magnetic field at the centre of the circular loop to the magnetic field at a distance $r$ from the centre of the loop on its axis is:

The magnetic field at the point of intersection of diagonals of a square wire loop of side $L$ carrying a current $I$ is

$A$ conducting wire carrying current $I$ is arranged as shown in the figure. Find the magnetic field at point $O$.

$A$ 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 $A$)

An electron revolves in a circle of radius $0.4 \text{ Å}$ with a speed of $10^6 \text{ m/s}$ in a hydrogen atom. The magnetic field produced at the centre of the orbit due to the motion of the electron (in Tesla) is: $\left[\mu_0 = 4\pi \times 10^{-7} \text{ H/m}, q = 1.6 \times 10^{-19} \text{ C}\right]$

Two concentric circular coils of $10$ turns each are situated in the same plane. Their radii are $20 \, cm$ and $40 \, cm$ and they carry respectively $0.2 \, A$ and $0.3 \, A$ current in opposite directions. The magnetic field at the centre is $(\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A)$