Question

Two wires AO and OC carry equal currents l as shown I figure. One end of both the wire extends to infinity. Angle AOC is $\alpha$. The magnitude of magnetic field at a point P on the bisector of these two wires at a distance r from point O is?

• A. $\frac{{\mu }_0}{2\pi }\frac{l}{r}\cot \left(\frac{\alpha }{2}\right)$
• B. $\frac{{\mu }_0}{4\pi }\frac{l}{r}\cot \left(\frac{\alpha }{2}\right)$
• C. $\frac{{\mu }_0}{2\pi }\frac{l}{r}\frac{\left(1+\cos \frac{\alpha }{2}\right)}{\sin \left(\frac{\alpha }{2}\right)}$
• D. $\frac{{\mu }_0}{4\pi }\frac{l}{r}\sin \left(\frac{\alpha }{2}\right)$

Answer 1

The correct option is C $\frac{{\mu }_0}{2\pi }\frac{l}{r}\frac{\left(1+\cos \frac{\alpha }{2}\right)}{\sin \left(\frac{\alpha }{2}\right)}$

$x=r\sin \frac{\alpha }{2}$

Magnetic field of point P

Due to OA and BC is in outward direction.

$B_P=\frac{{\mu }_{o}i}{4\pi x}(\cos \frac{\alpha }{2}-\cos \pi )+\frac{{\mu }_{o}i}{4\pi x}(\cos \frac{\alpha }{2}-\cos \pi )=\frac{{\mu }_{o}i}{4\pi x}(1+\cos \frac{\alpha }{2})+\frac{{\mu }_{o}i}{2\pi x}(1+\cos \frac{\alpha }{2})=\frac{{\mu }_{o}i}{2\pi x}(1+\cos \frac{\alpha }{2})=\frac{{\mu }_{o}i}{2\pi r}\frac{1+\cos \frac{\alpha }{2}}{\sin \frac{\alpha }{2}}$