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

Step 1: The bent wire is made of two semi-infinite straight arms meeting at the vertex, with angle \(\alpha\) between them. The bisector makes angle \(\alpha/2\) with each arm. The point \(P\) is on the bisector at distance \(x\) from the vertex.

Step 2: Perpendicular distance from \(P\) to each arm:
\[d = x\sin\left(\frac{\alpha}{2}\right)\]
Step 3: For a semi-infinite wire, \(B = \dfrac{\mu_0 i}{4\pi d}(\sin\theta_1 + \sin\theta_2)\). The far end (infinity) gives \(\sin\theta = 1\). The vertex end subtends an angle whose sine equals \(\cos(\alpha/2)\) (from the right triangle with legs \(x\sin\tfrac{\alpha}{2}\) and \(x\cos\tfrac{\alpha}{2}\)). Hence for one arm:
\[B_{arm} = \frac{\mu_0 i}{4\pi\, x\sin(\alpha/2)}\left(1 + \cos\frac{\alpha}{2}\right)\]
Step 4: Both arms give fields in the same sense at \(P\), so add them:
\[B = 2B_{arm} = \frac{\mu_0 i}{2\pi x}\cdot\frac{1+\cos(\alpha/2)}{\sin(\alpha/2)}\]
Step 5: Use \(1+\cos\phi = 2\cos^2\tfrac{\phi}{2}\) and \(\sin\phi = 2\sin\tfrac{\phi}{2}\cos\tfrac{\phi}{2}\) with \(\phi=\alpha/2\):
\[\frac{1+\cos(\alpha/2)}{\sin(\alpha/2)} = \cot\left(\frac{\alpha}{4}\right)\]
\[\boxed{B = \frac{\mu_0 i}{2\pi x}\cot\left(\frac{\alpha}{4}\right)}\]