Question bank/JEE Main/Physics

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

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

Show answer and solution▾

$\frac{\mu_{0}i}{4\pi R}\lbrack sin90^{\circ} - sin45^{\circ}\rbrack \otimes + \frac{\mu_{0}I}{2R} \odot + \frac{\mu_{0}i}{4\pi R}(sin45^{\circ} + sin90^{\circ}) \odot$
$= \frac{- \mu_{0}i}{4\pi R}\lbrack 1 - \frac{1}{\sqrt{2}}\rbrack + \frac{\mu_{0}i}{2R} + \frac{\mu_{0}i}{4\pi R}\lbrack\frac{1}{\sqrt{2}} + 1\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack - 1 + \frac{1}{\sqrt{2}} + 2\pi + \frac{1}{\sqrt{2}} + 1\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack\sqrt{2} + 2\pi\rbrack \odot$
$= \frac{\mu_{0}i}{2\pi R}\lbrack\frac{1}{\sqrt{2}} + \pi\rbrack \odot$

Source: Grass-G / P-19 Physics paper 28 Nov Fourth.docx