Question

A long wire carrying a current \(i\) is bent to form a plane angle \(\alpha\). The magnetic field \(B\) at a point on the bisector of this angle situated at a distance \(x\) from the vertex is:

• A. \(\dfrac{\mu_0 i}{2\pi x}\cot\left(\dfrac{\alpha}{4}\right)\)
• B. \(\dfrac{\mu_0 i}{2\pi x}\cos\left(\dfrac{\alpha}{4}\right)\)
• C. \(\dfrac{\mu_0 i}{2\pi x}\tan\left(\dfrac{\alpha}{4}\right)\)
• D. \(\dfrac{\mu_0 i}{2\pi x}\cot(\alpha)\)

Hint:

Each semi-infinite arm gives \(\tfrac{\mu_0 i}{4\pi d}(1+\cos\tfrac{\alpha}{2})\) with \(d = x\sin\tfrac{\alpha}{2}\); add the two and simplify with the half-angle identity.

Answer 1

The Correct Option is A

Setup: Treat the kink as two identical semi-infinite conductors sharing the vertex. By symmetry the point $P$ on the bisector is equidistant from both, at perpendicular distance $d = x\sin(\alpha/2)$, and each arm contributes an equal field in the same direction.

Field of one arm: Using $B = \dfrac{\mu_0 i}{4\pi d}(\sin\theta_1+\sin\theta_2)$ with the outer end at infinity ($\sin\theta_2 = 1$) and the vertex end giving $\sin\theta_1 = \cos(\alpha/2)$:
\[B_1 = \frac{\mu_0 i}{4\pi x\sin(\alpha/2)}\big(1+\cos(\alpha/2)\big)\]
Total: Doubling for the two arms,
\[B = \frac{\mu_0 i}{2\pi x}\cdot\frac{1+\cos(\alpha/2)}{\sin(\alpha/2)}\]
Applying the half-angle identity $\dfrac{1+\cos\phi}{\sin\phi}=\cot\dfrac{\phi}{2}$ with $\phi=\alpha/2$ collapses the bracket to $\cot(\alpha/4)$:
\[\boxed{B = \frac{\mu_0 i}{2\pi x}\cot\left(\frac{\alpha}{4}\right)}\]