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A steady current $I$ flows through a wire with one end at $O$ and the other end extending upto 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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Verified Answer
The correct answer is:
$\frac{\mu_0 I}{4 \pi d \sin \alpha}(1-\cos \alpha)$
From result of field of a long wire, we have
$$
\beta=\frac{\mu_0}{4 \pi} \times \frac{I}{r} \times\left(\sin \theta_1-\sin \theta_2\right)
$$
Here,
$$
\begin{aligned}
r & =d \sin \alpha \\
\sin \theta_1 & =\sin 90^{\circ}=1 \\
\sin \theta_2 & =\sin (90-\alpha)=\cos \alpha
\end{aligned}
$$
So, $\quad \beta=\frac{\mu_0 I}{4 \pi d \sin \alpha}(1-\cos \alpha)$
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