Question

Figure shows a very long wire ABDMNDC 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. $\frac{{\mu }_{0}I}{2\pi R}\left(\pi +\frac{1}{\sqrt{2}}\right)$
• B. $\frac{{\mu }_{0}I}{2R}$
• C. $\frac{{\mu }_{0}I}{2\pi R}\left(\pi -\frac{1}{\sqrt{2}}\right)$
• D. $\frac{{\mu }_{0}I}{2\pi R}(\pi +1)$

Answer 1

Answer: (A)

${\left(\overrightarrow{B}\right)}_0={\left({\left(\overrightarrow{B}\right)}_0\right)}_1+{\left({\left(\overrightarrow{B}\right)}_0\right)}_2+{\left({\left(\overrightarrow{B}\right)}_0\right)}_3+{\left({\left(\overrightarrow{B}\right)}_0\right)}_4$
${\overrightarrow{B}}_0={\left({\overrightarrow{B}}_0\right)}_1+{\left({\overrightarrow{B}}_0\right)}_2+{\left({\overrightarrow{B}}_0\right)}_3+{\left({\overrightarrow{B}}_0\right)}_4$
$\frac{{\mu }_{0}i}{4\pi R}\left[\sin 90^{\circ }-\sin 45^{\circ }\right]\otimes +\frac{{\mu }_{0}i}{2R}\odot +\frac{{\mu }_{0}i}{4\pi R}\left(\sin 45^{\circ }+\sin 90^{\circ }\right)\odot$
$=\frac{-{\mu }_{0}i}{4\pi R}\left[1-\frac{1}{\sqrt{2}}\right]+\frac{{\mu }_{0}i}{2R}+\frac{{\mu }_{0}i}{4\pi R}\left[\frac{1}{\sqrt{2}}+1\right]\odot$
$=\frac{{\mu }_{0}i}{4\pi R}\left[-1+\frac{1}{\sqrt{2}}+2\pi +\frac{1}{\sqrt{2}}+1\right]\odot =\frac{{\mu }_{0}i}{4\pi R}[\sqrt{2}+2\pi ]\odot =\frac{{\mu }_{0}i}{2\pi R}\left[\frac{1}{\sqrt{2}}+\pi \right]\odot$