Question

The magnetic field strength at $O$ due to current $I$ in the shown figure is :

• A. $\frac{7{\mu }_{0}I}{16R}$
• B. $\frac{15{\mu }_{0}I}{16R}$
• C. $\frac{11{\mu }_{0}I}{32R}$
• D. $\frac{13{\mu }_{0}I}{32R}$

Answer 1

The correct option is A $\frac{7{\mu }_{0}I}{16R}$

From Biot Savart's law
$\overrightarrow{B}=\int i\frac{(\overrightarrow{dl}\times \overrightarrow{r})}{r^3}$
Magnetic field at center $O$ due to portions of the loop $af$ and $cd$ will be zero since $\overrightarrow{dl}\parallel \overrightarrow{r}$ or $\overrightarrow{dl}\nparallel \overrightarrow{r}$ ( anti-parallel)
Magnetic field at the center of a circular coil is $B=\frac{{\mu }_{0}i}{2r}$
Magnetic field at center $O$ due to portion of the $abc$ is:
$B_1=\frac{3}{4}\frac{{\mu }_{0}i}{2R}$ since portion $abc$ is ${\frac{3}{4}}^{th}$ of the circular loop
Magnetic field at center $O$ due to portion of the $cdef$ is: $B_2=\frac{1}{4}\frac{{\mu }_{0}i}{4R}$ since portion $cdef$ is ${\frac{1}{4}}^{th}$ of the circular loop
$B=B_1+B_2=\frac{3}{4}\times \frac{{\mu }_{0}i}{2R}+\frac{1}{4}\times \frac{{\mu }_{0}i}{4R}=\frac{7{\mu }_{0}i}{16R}$ since both $B_1$ and $B_2$ point in the same direction.