Match List I with List II List – I (Current configuration) List –...

(A)

Here net magnetic field will be sum of magnetic field due to straight wires $ab, de$ and due to loop $bcd$.

So, ${B}_{ab} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} = {B}_{de}$ (directed out of the plane in both cases)

${B}_{bcd} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( 2\pi )$ (in the plane)

Thus, magnetic field at $O$ is

${B}_{O} = - \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} + \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( 2\pi ) - \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r}$

${B}_{O} = \dfrac{{\mu}_{0}}{2\pi}\dfrac{I}{r} ( \pi - 1 )$ (III)

(B)

Here net magnetic field will be sum of magnetic field due to straight wires $ab, de$ and semicircular arc $bcd$.

${B}_{ab} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} = {B}_{de}$ (directed out of the plane in both cases)

${B}_{bcd} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( \pi )$ (out of the plane)

Thus, net magnetic field at $O$ is

${B}_{O} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} + \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( \pi ) + \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r}$

${B}_{0} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( \pi + 2 )$ (I)

(C)

Here net magnetic field will be sum of magnetic field due to straight wires $ab, de$ and due to loop $bcd$.

${B}_{ab} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r}$ (directed in the plane)

${B}_{bcd} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( \pi )$ (in the plane)

${B}_{de} = 0$ (passing through the axis)

Thus, magnetic field at $O$ is

${B}_{O} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( \pi + 1 )$ (IV)

(D)

Here net magnetic field will be sum of magnetic field due to straight wires $ab, de$ and due to loop $bcd$.

${B}_{ab} = 0 = {B}_{de}$ (passing through the axis)

${B}_{bcd} = \dfrac{{\mu}_{0}}{4\pi}\dfrac{I}{r} ( \pi )$ (out of the plane)

Thus, magnetic field at $O$ is

${B}_{O} = \dfrac{{\mu}_{0}I}{4r}$ (II).

A-III, B-I, C-IV, D-II