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. $\frac{\mu _{0} I}{2 R}$
• 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)

$\left(\overset{ \rightarrow }{B}\right)_{0}=\left(\left(\overset{ \rightarrow }{B}\right)_{0}\right)_{1}+\left(\left(\overset{ \rightarrow }{B}\right)_{0}\right)_{2}+\left(\left(\overset{ \rightarrow }{B}\right)_{0}\right)_{3}+\left(\left(\overset{ \rightarrow }{B}\right)_{0}\right)_{4}$
$\overrightarrow{ B }_{0}=\left(\overrightarrow{ B }_{0}\right)_{1}+\left(\overrightarrow{ B }_{0}\right)_{2}+\left(\overrightarrow{ B }_{0}\right)_{3}+\left(\overrightarrow{ B }_{0}\right)_{4}$
$\frac{\mu_{0} i }{4 \pi R }\left[\sin 90^{\circ}-\sin 45^{\circ}\right] \otimes+\frac{\mu_{0} i }{2 R } \odot+\frac{\mu_{0} i }{4 \pi R }\left(\sin 45^{\circ}+\sin 90^{\circ}\right) \odot$
$=\frac{-\mu_{0} i }{4 \pi R }\left[1-\frac{1}{\sqrt{2}}\right]+\frac{\mu_{0} i }{2 R }+\frac{\mu_{0} i }{4 \pi R }\left[\frac{1}{\sqrt{2}}+1\right] \odot$
$=\frac{\mu_{0} i }{4 \pi R }\left[-1+\frac{1}{\sqrt{2}}+2 \pi+\frac{1}{\sqrt{2}}+1\right] \odot \quad=\frac{\mu_{0} i }{4 \pi R }[\sqrt{2}+2 \pi] \odot=\frac{\mu_{0} i }{2 \pi R }\left[\frac{1}{\sqrt{2}}+\pi\right] \odot$