Question

A long wire carrying a current i is bent to form a place along α . Find the magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex.

Answer 1

Let CAB be the wire making an angle α, P be the point on the bisector of this angle situated at a distance x from the vertex A and d be the perpendicular distance of AC and AB from P. 

From the figure,

\[\sin \left( \frac{\alpha}{2} \right) = \frac{d}{x}\]

\[d   =   x\sin  \left( \frac{\alpha}{2} \right)\]

The angles made by points A and C with point P are

\[\theta_1 = 90 - \frac{\alpha}{2} \text{ and } \theta_2 = 90^\circ \] , respectively.

Separation of the point from the wire, \[d   =   x\sin  \left( \frac{\alpha}{2} \right)\]

 Thus, the magnetic field due to current in wire AC is given by 

\[B = \frac{\mu_0 i}{4\pi d}(\sin \theta_1 + \sin \theta_2 )\]
\[ = \frac{\mu_0 i}{4\pi x\sin \frac{\alpha}{2}}\left[ \sin \left( 90 - \frac{\alpha}{2} \right) + \sin 90 \right]\]

\[\frac{\mu_0 i}{4\pi x\sin  \left( \frac{\alpha}{2} \right)}\left[ \cos  \frac{\alpha}{2} + 1 \right]\] 

\[ = \frac{\mu_0 i2 \cos^2 \left( \frac{\alpha}{4} \right)}{4\pi x2\sin  \left( \frac{\alpha}{4} \right)\cos  \left( \frac{\alpha}{4} \right)}   =   \frac{\mu_0 i\cot  \left( \frac{\alpha}{4} \right)}{4\pi x}\]

Now, the magnetic field due to wires AC and AB is given by

\[B_{net} = 2B = \frac{\mu_0 i\cot \left( \frac{\alpha}{4} \right)}{2\pi x}\]