Current through ABC and A'B'C' is I as shown in the given figure. If, $PB = PB'=r$ and $C'B'PBC$ are collinear, the magnetic field at P is:

$2μ_0I/4πг$

The correct answer is Option (2) → $2μ_0I/4πг$

Magnetic field due to the portions BC and B'C' are zero.

Magnetic field due to AB:

$B_1=\frac{μ_0I}{4π(PB)}[\sin θ_1+\sin θ_2]$

$=\frac{μ_0I}{4πr}[0+1]$   ($θ_1=0°$ and $θ_2=90°$)

$=\frac{μ_0I}{4πr}$

Shy, Magnetic field due to A'B':

$B_2=\frac{μ_0I}{4πr}$

$∴B_1+B_2=\frac{μ_0I(2)}{4πr}$