Magnetic field due to the current carrying wire as shown in the figure at point "O" will be :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{\mu_0 I}{4R} \)
The wire has three parts: the straight part on the left, the semicircle of radius \( R \) centred at O, and the straight part on the right.
The two straight parts lie on the line through O, so for every element of them \( d\vec{l} \) is parallel to \( \vec{r} \) and, by the Biot-Savart law, \( d\vec{l}\times\vec{r} = 0 \). They give no field at O.
A full circular loop gives \( \mu_0 I/2R \) at its centre. A semicircle is half the loop, and every element of it gives a field in the same direction at O, so
\( \displaystyle B = \frac{1}{2}\times\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4R} \)