Question

A long wire carrying current $i$ has been bent into the shape as shown in figure. $AB$ and $CD$ are circular arcs centred at $O$, having radius $2r$ and $3r$ respectively. $EF$ is a semicircle of radius $r$ centred at $O$, all other segments are straight and radial pointing towards $O$. Find the magnetic field at the point $O$.

• A. $\frac{{\mu }_{0}i}{35r}$
• B. $\frac{144{\mu }_{0}i}{35r}$
• C. $\frac{35{\mu }_{0}i}{144r}$
• D. $\frac{25{\mu }_{0}i}{144r}$

Answer 1

The correct option is C $\frac{35{\mu }_{0}i}{144r}$

Field due to all straight wire segment at $O$ is zero.

Field due to the circular arcs are:
$B_{AB}=\frac{{\mu }_{0}I}{2\left[3r\right]}\left[\frac{1}{12}\right]=\frac{{\mu }_{0}I}{72r}\otimes B_{CD}=\frac{{\mu }_{0}I}{2\left[2r\right]}\left[\frac{1}{12}\right]=\frac{{\mu }_{0}I}{48r}\odot B_{EF}=\frac{{\mu }_{0}I}{4r}\otimes \text{resultant field is,}B=(\frac{{\mu }_{0}I}{72r}+\frac{{\mu }_{0}I}{4r}-\frac{{\mu }_{0}I}{48r})\otimes \Rightarrow B=\frac{35}{144}\frac{{\mu }_{0}I}{r}\otimes$