$A$ coil having $9$ turns carrying a current produces a magnetic field $B_1$ at the centre. Now,the coil is rewound into $3$ turns carrying the same current. Then,the magnetic field at the centre $B_2$ is:

For a wire, as shown in the figure, carrying a current of $10 \,A$, find the magnetic induction field at the point $O$. (Given: $\mu_0 = 4 \pi \times 10^{-7} \,H/m$)

Two coils $P$ and $Q$ each of radius $R$ carry currents $I$ and $\sqrt{8} I$ respectively in the same direction. These coils are lying in perpendicular planes such that they have a common centre. The magnitude of the magnetic field at the common centre of the two coils is ($\mu_0 =$ permeability of free space).

The magnitude of the magnetic field at $O$ due to a current-carrying loop as shown in the figure is, where $O$ is the center of two circular portions with radii $1 \, cm$ and $2 \, cm$ respectively. (Take the value of current $I = \frac{1.2}{\pi} \, A$)

The $SI$ unit of magnetic permeability is

Two long parallel wires carrying currents $I_1 = 4 \ A$ and $I_2 = 3 \ A$ in opposite directions are placed at a distance of $d = 5 \ cm$ from each other. $A$ point $P$ is at equidistance from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of the magnetic field at point $P$ is ( $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$ ).