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

Match List I with List II List – I (Current… - JEE Main 2023 Physics PYQ

JEE Main 2023 — Physics Electromagnetism

2023mcqhard

Match List I with List II

List – I (Current configuration)	List – II (Magnetic field at point $O$ )
A	I.	${B}_{0}=\frac{{\mu }_{0}I}{4\pi r}[\pi +2]$
B	II.	${B}_{0}=\frac{{\mu }_{0}}{4}\frac{I}{r}$
C	III.	${B}_{0}=\frac{{\mu }_{0}I}{2\pi r}[\pi -1]$
D	IV.	${B}_{0}=\frac{{\mu }_{0}I}{4\pi r}[\pi +1]$

Choose the correct answer from the option given below:

Official previous-year question

Held on 25 Jan 2023 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

(A)

JEE Main 2023 Physics, Magnetic Effects of Current — solution figure 1

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

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

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

Thus, magnetic field at $O$ is

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

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

(B)

JEE Main 2023 Physics, Magnetic Effects of Current — solution figure 2

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

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

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

Thus, net magnetic field at $O$ is

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

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

(C)

JEE Main 2023 Physics, Magnetic Effects of Current — solution figure 3

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

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

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

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

Thus, magnetic field at $O$ is

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

(D)

JEE Main 2023 Physics, Magnetic Effects of Current — solution figure 4

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

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

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

Thus, magnetic field at $O$ is

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

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

Your note

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