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).

$A$ steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns, the magnetic field at the centre of the coil is

Two concentric coplanar circular loops of radii $r_1$ and $r_2$ carry currents $i_1$ and $i_2$ in opposite directions (one clockwise and the other anticlockwise). The magnetic induction at the center of the loops is half that due to $i_1$ alone at the center. If $r_2 = 2r_1$,find the value of $\frac{i_2}{i_1}$.

$A$ straight wire of length $20 \text{ cm}$ carrying a current of $\frac{3}{\pi^2} \text{ A}$ is bent in the form of a circle. The magnetic field at the centre of the circle is

Two concentric coils of $10$ turns each are placed in the same plane. Their radii are $20 \ cm$ and $40 \ cm$ and carry $0.2 \ A$ and $0.3 \ A$ current respectively in opposite directions. The magnetic induction (in $T$) at the centre is

$A$ 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 $(b)$?