Magnetic field |B| at a point P in the following network is:

The correct answer is Option (3) → \( \displaystyle\frac{\mu_0 I}{4\pi R}\left(\frac{3}{2}\pi - 2\right) \)

The field at P is the vector sum, by the Biot–Savart law, of the field of the circular arc and the fields of the two straight arms.

The arc is three-quarters of a circle of radius R and has P at its centre. A full circle gives \( \frac{\mu_0 I}{2R} \) at the centre, so

    \( \displaystyle B_{\mathrm{arc}} = \frac{3}{4}\times\frac{\mu_0 I}{2R} = \frac{\mu_0 I}{4\pi R}\left(\frac{3\pi}{2}\right) \)

Each straight arm lies along the tangent at an end of the arc, so the perpendicular from P meets it exactly at its end, at distance R. A straight wire seen from the foot of the perpendicular at its end is half of an infinite wire:

    \( \displaystyle B_{\mathrm{arm}} = \frac{1}{2}\times\frac{\mu_0 I}{2\pi R} = \frac{\mu_0 I}{4\pi R} \)

The current in the two arms flows in the sense (see figure) that makes their field at P opposite to that of the arc, so it is subtracted:

    \( \displaystyle B = B_{\mathrm{arc}} - 2B_{\mathrm{arm}} = \frac{\mu_0 I}{4\pi R}\left(\frac{3\pi}{2} - 2\right) \)