Question 53 of 276

Match List I with List II List – I
(Current configuration) List – II
(Magnetic field at point $O$) A I. ${B}_{0}=\dfrac{{\mu }_{0}I}{4\pi r}[\pi +2]$ B II. ${B}_{0}=\dfrac{{\mu }_{0}}{4}\dfrac{I}{r}$ C III. ${B}_{0}=\dfrac{{\mu }_{0}I}{2\pi r}[\pi -1]$ D IV. ${B}_{0}=\dfrac{{\mu }_{0}I}{4\pi r}[\pi +1]$ Choose the correct answer from the option given below:

AA-III, B-IV, C-I, D-IIBA-I, B-III, C-IV, D-IICA-III, B-I, C-IV, D-IIDA-II, B-I, C-IV, D-III

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Answer: (C) A-III, B-I, C-IV, D-II

(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