Question 75 of 276

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 filed at the center of circle is:

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

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Answer: (A) $\dfrac{{\mu }_{0}I}{2\pi R}\left(\pi +\dfrac{1}{\sqrt{2}}\right)$

$\vec{{B}_{0}}={\left(\vec{{B}_{0}}\right)}_{1}+{\left(\vec{{B}_{0}}\right)}_{2}+{\left(\vec{{B}_{0}}\right)}_{3}+{\left(\vec{{B}_{0}}\right)}_{4}$
$\dfrac{{\mu }_{0}i}{4\pi R}\left[\sin ⁡90^{\circ}-sin45^{\circ}\right]⨂+\dfrac{{\mu }_{0}i}{2R}⨀+$ $\dfrac{{\mu }_{0}i}{4\pi R}(sin45^{\circ}+sin90^{\circ})⨀$
$-\dfrac{{\mu }_{0}i}{4\pi R}\left[1-\dfrac{1}{\sqrt{2}}\right]+\dfrac{{\mu }_{0}i}{2R}+\dfrac{{\mu }_{0}i}{4\pi R}\left[\dfrac{1}{\sqrt{2}}+1\right]⨀$
$=\dfrac{{\mu }_{0}i}{4\pi R}\left[-1+\dfrac{1}{\sqrt{2}}+2\pi +\dfrac{1}{\sqrt{2}}+1\right]⨀=\dfrac{{\mu }_{0}i}{4\pi R}\left[\sqrt{2}+2\pi \right]⨀=\dfrac{{\mu }_{0}i}{2\pi R}\left[\dfrac{1}{\sqrt{2}}+\pi \right]⨀$