Match the List-I with List- II :

List- I
(Current configuration)

List-II
(Magnitude of Magnetic field at point O)
A)
I) $B_{0} = \frac{μ_{0} I}{4 πr} [π + 2]$
B)
II) $B_{0} = \frac{μ_{0}}{4} \frac{I}{r}$
C)
III) $B_{0} = \frac{μ_{0} I}{2 πr} [π - 1]$
D)
IV) $B_{0} = \frac{μ_{0} I}{4 πr} [π + 1]$
Choose the correct answer from the options given below :

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A – I; B – III; C – IV; D – II

A – III; B – I; C – IV; D – II

A – II; B – I; C – IV; D – III

A – III; B – IV; C – I; D – II

answer is D.

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Detailed Solution

$\begin{array}{l} A \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ + \frac{μ_{0} i}{2 r} \otimes + \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ \\ = \frac{μ_{0}}{2 π} \frac{i}{r} ⊙ + \frac{μ_{0} i}{2 r} \otimes \\ B_{0} = \frac{μ_{0} i}{2 rπ} (π - 1) \otimes \\ B \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ + \frac{μ_{0}}{4 π} \frac{i}{r} π ⊙ + \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ \\ = (\frac{2 μ_{0}}{4 π} \frac{i}{r} + \frac{μ_{0}}{4 π} \frac{i}{r} π) ⊙ \\ B_{0} = \frac{μ_{0} i}{4 πr} (2 + π) ⊙ \\ C \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} \otimes + \frac{μ_{0}}{4 π} \frac{i}{r} π \otimes + 0 \\ B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} (1 + π) \otimes \\ D \to B_{0} = \frac{μ_{0}}{4 π} \frac{i}{r} π = \frac{μ_{0}}{4 π} \frac{i}{r} ⊙ \end{array}$

A – III; B – I; C – IV; D – II

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