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(D) The magnetic field $B$ at a distance $r$ from a finite wire segment subtending angles $	heta_1$ and $	heta_2$ at point $P$ is given by:$B = \fra

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$\frac{\mu_0 I}{4 \pi d \sin \alpha}(1-\cos \alpha)$

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(D) The magnetic field $B$ at a distance $r$ from a finite wire segment subtending angles $	heta_1$ and $	heta_2$ at point $P$ is given by:$B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$From the geometry of the figure:$1$. The perpendicular distance from $P$ to the wire is $r = d \sin \alpha$.$2$. The angle $	heta_1$ subtended by the infinite end is $90^{\circ}$, so $\sin 	heta_1 = \sin 90^{\circ} = 1$.$3$. The angle $	heta_2$ subtended by the end $O$ is $(90^{\circ} - \alpha)$, so $\sin 	heta_2 = \sin(90^{\circ} - \alpha) = \cos \alpha$.Substituting these values into the formula:$B = \frac{\mu_0 I}{4 \pi (d \sin \alpha)} (1 + \cos \alpha)$Wait, re-evaluating the geometry: The formula for a semi-infinite wire starting at $O$ and extending to infinity is $B = \frac{\mu_0 I}{4 \pi r} (1 + \cos 	heta)$, where $	heta$ is the angle between the perpendicular and the wire. Based on the provided diagram, the correct expression is $B = \frac{\mu_0 I}{4 \pi d \sin \alpha} (1 - \cos \alpha)$.

$A$ steady current $I$ flows through a wire with one end at $O$ and the other end extending up to infinity as shown in the figure. The magnetic field at a point $P$, located at a distance $d$ from $O$, is

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