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?

$A$ straight wire carrying a current of $12\; A$ is bent into a semi-circular arc of radius $2.0\; cm$ as shown in Figure $(a)$. Consider the magnetic field $B$ at the centre of the arc.
$(a)$ What is the magnetic field due to the straight segments?
$(b)$ In what way does the contribution to $B$ from the semicircle differ from that of a circular loop and in what way does it resemble?
$(c)$ Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Figure $(b)$?

$A$ long current-carrying wire produces a magnetic field of $1 \ T$ at a distance of $r$. What will be the magnetic field at distances of $(a)$ $\frac{r}{2}$,$(b)$ $2r$,and $(c)$ $3r$?

$A$ 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

$A$ current of $1 \,A$ is flowing along the sides of an equilateral triangle of side $a = 4.5 \times 10^{-2} \,m$. The magnetic field at the centroid of the triangle is $(\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A)$.

The magnetic field due to a current-carrying circular loop of radius $6 \text{ cm}$ at a point on the axis at a distance of $8 \text{ cm}$ from the centre is $27 \mu \text{T}$. The magnetic field at the centre of the current-carrying loop is: (in $\mu \text{T}$)