Question

The magnetic induction at point $O$ in the following figure will be -

• A. $\frac{{\mu }_{0}I}{4r}[\frac{3}{2}+\frac{1}{\pi }]$ $⨂$
  
• B. $\frac{{\mu }_{0}I}{4r}[\frac{3}{2}+\frac{1}{\pi }]$ $⨀$
• C. $\frac{{\mu }_{0}I}{4r}[\frac{3}{2}-\frac{1}{\pi }]$ $⨂$
• D. $\frac{{\mu }_{0}I}{4r}[\frac{3}{2}-\frac{1}{\pi }]$ $⨀$

Answer 1

The correct option is C $\frac{{\mu }_{0}I}{4r}[\frac{3}{2}-\frac{1}{\pi }]$ $⨂$


Net magnetic field,

$B=B_1+B_2+B_3...\left(1\right)$

$B_1=0...\left(2\right)$

$B_2=\frac{{\mu }_{0}I}{2r}\frac{\theta }{2\pi }=\frac{{\mu }_{0}I}{2r}\times \frac{3\pi /2}{2\pi }$

$\Rightarrow B_2=\frac{3{\mu }_{0}I}{8r}$ $⨂...\left(3\right)$

$B_3=\frac{{\mu }_{0}I}{4\pi r}$$[\sin {\theta }_1+\sin {\theta }_2]$

$=\frac{{\mu }_{0}I}{4\pi r}$ $[\sin 0^{\circ }+\sin {90}^{\circ }]$

$=\frac{{\mu }_{0}I}{4\pi r}⨀...\left(4\right)$

From $\left(1\right),\left(2\right),\left(3\right)$ and $\left(4\right)$,

$B=0+[\frac{3}{2}-\frac{1}{\pi }]\frac{{\mu }_{0}I}{4r}$

$\Rightarrow B=\frac{{\mu }_{0}I}{4r}$ $[\frac{3}{2}-\frac{1}{\pi }]$ $⨂$