A straight wire carrying a current of $12 \mathrm{~A}$ is bent into a semi-circular arc of radius $2…
A straight wire carrying a current of $12 \mathrm{~A}$ is bent into a semi-circular arc of radius $2 \mathrm{~cm}$ as shown in the figure. Then the magnetic field due to the straight segments at the centre of the arc is
$12 \mathrm{~T}$
$6 \mathrm{~T}$
$24 \mathrm{~T}$
0
Solution
(dFor a point $\mathrm{O}$ d $\overrightarrow{\mathrm{l}}$ and $\overrightarrow{\mathrm{r}}$ for each element of the straigh segment $\mathrm{AB}$ and $\mathrm{DE}$ are parallel. Therefor, $\mathrm{d} \overrightarrow{\mathrm{l}} \times \overrightarrow{\mathrm{r}}=0$
Hence, magnetic field at $\mathrm{O}$ due to current through straight segments $\mathrm{AB}$ and $\mathrm{DE}$ is
$
\overrightarrow{\mathrm{B}}=\int \frac{\mu_0}{4 \pi} \frac{\mathrm{id} \overrightarrow{\mathrm{l}} \times \overrightarrow{\mathrm{r}}}{\mathrm{r}^3}=0
$
