Question

A long straight wire, carrying current I, is bent at its midpoint to from an angle of ${45}^{\circ }$ Induction of magnetic field at point P, distant R from point of bending is equal to :

• A. $\frac{(\sqrt{2}-1){\mu }_{0}I}{4\pi R}$
• B. $\frac{(\sqrt{2}+1){\mu }_{0}I}{4\pi R}$
• C. $\frac{(\sqrt{2}-1{)}_{10}1}{4\sqrt{2}\pi R}$
• D. $\frac{(\sqrt{2}+1){\mu }_{0}1}{4\sqrt{2}\pi R}$

Answer 1

The correct option is A $\frac{(\sqrt{2}-1){\mu }_{0}I}{4\pi R}$

$\begin{array}{c}Accordingtoquestion..............\\ thepointliesontheaxisofstraightpartOA,\\ MagneticfieldatpointPiszeroduetoapartOAofwire.\\ and,forpartOC,from\Delta OPN,\\ \therefore d=R\cos {45}^0\\ sinceboththeendsO\&CareonthesamesideofthenormalPN,\\ So,\\ {\varphi }_1=-{45}^{0}and\varphi ={90}^0\\ B=\frac{{\mu }_{0}i}{4\pi d}\left[\sin {\varphi }_1+\sin {\varphi }_2\right]\\ \Rightarrow B=\frac{{\mu }_{0}i}{4\pi R\cos {45}^0}\left[\sin (-{45}^0)+\sin \left({90}^0\right)\right]\\ \Rightarrow B=\frac{{\mu }_{0}i\times \sqrt{2}}{4\pi R}\left[-\sin {45}^0+1\right]\\ \Rightarrow B=\frac{\sqrt{2}{\mu }_{0}i}{4\pi R}\left[1-\frac{1}{\sqrt{2}}\right]\\ \Rightarrow B=\frac{\sqrt{2}{\mu }_{0}i}{4\pi R}\left[\frac{\sqrt{2}-1}{\sqrt{2}}\right]\\ \therefore B=\frac{{\mu }_{0}i}{4\pi R}\left[\sqrt{2}-1\right]\\ B=\frac{\left[\sqrt{2}-1\right]{\mu }_{0}i}{4\pi R}\\ sothatthecorrectoptionisA.\end{array}$