Question bank/JEE Main/Physics

A very long wire ABDMNDC is shown in figure carrying current I.
AB and BC parts are straight, longand at right angle. At D wire forms a
circular turn DMND of radius R. AB, BC parts are tangential tocircular
turn at N and D. Magnetic field at the centre of circle is

A$\frac{\mu_{0}I}{2\pi R}\left( \pi + \frac{1}{\sqrt{2}} \right)$B$\frac{\mu_{0}I}{2R}$C$\frac{\mu_{0}I}{2\pi R}\left( \pi - \frac{1}{\sqrt{2}} \right)$D$\frac{\mu_{0}I}{2\pi R}(\pi + 1)$

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$\frac{\mu_{0}i}{4\pi R}\left\lbrack \sin 90^{\circ} - \sin 45^{\circ} \right\rbrack \otimes + \frac{\mu_{0}i}{2R} \odot + \frac{\mu_{0}i}{4\pi R}\left( \sin 45^{\circ} + \sin 90^{\circ} \right) \odot$
$= \frac{- \mu_{0}i}{4\pi R}\left\lbrack 1 - \frac{1}{\sqrt{2}} \right\rbrack + \frac{\mu_{0}j}{2R} + \frac{\mu_{0}i}{4\pi R}\left\lbrack \frac{1}{\sqrt{2}} + 1 \right\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\left\lbrack - 1 + \frac{1}{\sqrt{2}} + 2\pi + \frac{1}{\sqrt{2}} + 1 \right\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack\sqrt{2} + 2\pi\rbrack \odot$
$= \frac{\mu_{0}i}{2\pi R}\left\lbrack \frac{1}{\sqrt{2}} + \pi \right\rbrack \odot$

Source: Grass-G / P-13 Physics paper 4 Nov.docx