The magnetic field due to a current-carrying loop of radius $3 \text{ cm}$ at a point on its axis at a distance of $4 \text{ cm}$ from its centre is $54 \mu\text{T}$. Then,the value of the magnetic field at the centre of the loop is: (in $\mu\text{T}$)

For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

Two circular coils $1$ and $2$ are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$,so that the magnetic field at their centre is the same?

For the adjoining figure,the magnetic field at point $P$ will be:

The magnetic field at the centre $O$ of a wire loop formed by two semicircular wires of radii $R_1=2 \pi \text{ m}$ and $R_2=4 \pi \text{ m}$ carrying current $I=4 \text{ A}$ as per the figure given below is $\alpha \times 10^{-7} \text{ T}$. The value of $\alpha$ is . . . . . . . (Centre $O$ is common for all segments)

$A$ long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in the $X-Y$ plane and a steady current $I$ flows through the wire. The $Z$-component of the magnetic field at the center of the spiral is