$A$ portion of a conductive wire is bent in the form of a semicircle of radius $r$ as shown in the figure. At the centre $O$ of the semicircle,the magnetic induction will be:

The radius of a circular current-carrying coil is $R$. At what distance from the centre of the coil on its axis,the intensity of the magnetic field will be $\frac{1}{2 \sqrt{2}}$ times that at the centre?

$A$ length $L$ of wire carries a steady current $I$. It is bent first to form a circular plane coil of one turn. The same length is now bent more sharply to give a double loop of smaller radius. The magnetic field at the centre caused by the same current is

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,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

The magnetic field at the center of a circular coil of radius $0.1\, m$ having $1000$ turns and carrying a current of $0.1\, A$ is:

$A$ long straight wire carries a current of $35\, A$. What is the magnitude of the magnetic field $B$ at a point $20\, cm$ from the wire?