Question

A very long wire $ABDMNDC$ is shown in figure carrying current $I$. $AB$ and $BC$ parts are straight, long and at right angle. At $D$ the wire forms a circular turn $DMND$ of radius $R$. $AB$, $BC$ parts are tangential to the 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. $\frac{{\mu }_{0}I}{2\pi R}(\pi +1)$
  
• C. $\frac{{\mu }_{0}I}{2\pi R}(\pi -\frac{1}{\sqrt{2}})$
• D. $\frac{{\mu }_{0}I}{2R}$

Answer 1

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



Field due to a straight wire is given by
${\overrightarrow{B}}_0=\frac{{\mu }_{0}i}{2\pi }(\sin {\varphi }_1+\sin {\varphi }_2)$
($r$=distance of point from wire & ${\varphi }_1,{\varphi }_2$ are angle made by line joining two ends of wire with the given point with perpendicular on wire from given point )

At $O$, magnetic field of wire $1$ ,
$B_1=\frac{{\mu }_{0}i}{4\pi r}(\sin {90}^0+\sin -{45}^0)$
$B_1=\frac{{\mu }_{0}i}{4\pi r}(1-\frac{1}{\sqrt{2}})$
At $O$ , due to circular loop,
$B_3=\frac{{\mu }_{0}i}{2R}$ $⨀$
At $O$, due to wire $2$ and $4$ combined ,
$B_2+B_4=\frac{{\mu }_{0}i}{4\pi r}(\sin {45}^0+\sin {90}^0)$$⨀$
Net field at $O$,
$B_{net}=B_1+B_2+B_3+B_4$
$B_{net}=-\frac{{\mu }_{0}i}{4\pi r}[1-\frac{1}{\sqrt{2}}]+\frac{{\mu }_{0}i}{2R}+\frac{{\mu }_{0}i}{4\pi r}$$[\frac{1}{\sqrt{2}}+1]$ $⨀$
$B_{net}=\frac{{\mu }_{0}i}{4\pi r}[-1+\frac{1}{\sqrt{2}}+2\pi +\frac{1}{\sqrt{2}}+1]$ $⨀$
$B_{net}=\frac{{\mu }_{0}i}{4\pi r}[\sqrt{2}+2\pi ]$ $⨀$
$B_{net}=\frac{{\mu }_{0}i}{4\pi r}[\pi +\frac{1}{\sqrt{2}}]$ $⨀$
Hence, answer is $\left(C\right)$.