$A$ circular coil carrying current $I$ has a radius $r$ and $n$ turns. The magnetic field along the axis of a coil at a distance $x = 2\sqrt{2}r$ from its centre is (where $\mu_0$ is the permeability of free space).

When a helium nucleus makes a full rotation of a circle of radius $0.8 \,m$ in $2.5 \,s$, then the value of magnetic field $B$ at the centre of the circle will be

Describe Oersted's observation.

Magnetic lines of force due to a current in a circular conductor are

$A$ very long wire carrying a current $I = 4 \sqrt{2} \, A$ is bent at a right angle. The magnitude of the magnetic field $(|B|)$ at a point $P$ lying on a line perpendicular to the bent wire at a distance $d = 20 \, cm$ from the point of the bending will be (Let $\mu_0 = 4 \pi \times 10^{-7} \, H/m$): (in $\, \mu T$)

The current through $ABC$ and $A'B'C'$ is $I$. What is the magnetic field at $P$? Given $BP = PB' = r$ (Here $C', B', P, B, C$ are collinear).