Verified Solution • 2025–26 Session moderate

QJEEM23Ph5631: JEE Main 2023 Physics - Class 12 JEEM23Ph5631 Solution: Match List I with List II .tg…

Match List I with List II

	List I (Current configuration)		List II (Magnitude of Magnetic Field at point O)
A.		I.	$B_{0} = \frac{μ_{0} I}{4 π r} [π + 2]$
B.		II.	$B_{0} = \frac{μ_{0}}{4} \frac{I}{r}$
C.		III.	$B_{0} = \frac{μ_{0} I}{2 π r} [π - 1]$
D.		IV.	$B_{0} = \frac{μ_{0} I}{4 π r} [π + 1]$

Choose the correct answer from the options given below :

Multiple Choice Options

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

Step-by-Step Verified Solution

A \rightarrow B_{0}=\frac{-\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{2 r}-\frac{\mu_{0} I}{4 \pi r}

B_{0} = \frac{μ_{0} I}{2 π r} (π - 1) A \to III

B \rightarrow B_{0}=\frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}+\frac{\mu_{0} I}{4 \pi r}

B_{0} = \frac{μ_{0} I}{4 π r} (π + 2) B \to I

C \rightarrow B_{0}=\frac{\mu_{0} I}{4 \pi r}+\frac{\mu_{0} I}{4 r}+0

B_{0}=\frac{\mu_{0} I}{4 \pi r}(\pi+1) \quad \quad \mathrm{C} \rightarrow \mathrm{IV}

D.

\rightarrow B_{0}=\frac{\mu_{0} I}{4 r} \quad \quad \mathrm{D} \rightarrow \mathrm{II}

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