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

Explanation:

To find: The magnetic field B at a point on the bisector of this angle situated at a distance x from the vertex.

Step 1:

Let CAB be the wire making an angle α, let P be the point on this angle's bisector located at a distance x from the vertex A, and let d be the perpendicular distance between AC and AB and P.  From the representation,

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

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

Point A and point C coincides with point P

$\theta_{1}=90-\alpha_{2} \text { and } \theta_{2}=90^{\circ}$

Dividing the point from the wire

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

Step 2:

Thus the magnetic field is given by the current in wire AC

$B=\frac{\mu_{0} i}{4 \pi d}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

$=\frac{\mu_{0} i}{4 \pi x \sin \left(\frac{\alpha}{2}\right)}\left[\sin \left(90-\frac{\alpha}{2}\right)+\sin 90\right]$

$=\frac{\mu_{0} i}{4 \pi x \sin \left(\frac{\alpha}{2}\right)}\left[\cos \frac{\alpha}{2}+1\right]$

$=\frac{\mu_{0} i 2 \cos ^{2} \frac{\alpha}{4}}{4 \pi x 2 \sin \left(\frac{\alpha}{4}\right) \cos \frac{\alpha}{4}}$

$=\frac{\mu_{0} i \cot \frac{\alpha}{4}}{4 \pi x}$

Step 3:

The magnetic field attributable to AC and AB wires is now given by

$B_{\text {net }}=2 B=\frac{\mu_{0} i \cot \frac{\alpha}{4}}{2 \pi x}$