Question

A long straight wire, carrying current I, is bent at its midpoint to form an angle of ${45}^0$. Magnetic field at point P at a distance R from point of bending is equal to :

• A. $\frac{(\sqrt{2}-1){\mu }_{0}I}{4\pi R}$
• B. $\frac{(\sqrt{2}+1){\mu }_{0}I}{4\pi R}$
• C. $\frac{(\sqrt{2}+1){\mu }_{0}I}{4\sqrt{2}\pi R}$
• D. $\frac{(\sqrt{2}-1){\mu }_{0}I}{2\sqrt{2}\pi R}$

Answer 1

The correct option is A $\frac{(\sqrt{2}-1){\mu }_{0}I}{4\pi R}$

Since point P lies on the axis of the straight part OA , magnetic field at point P is zero due to wire OA .
For part OC , perpendicular distance of Point P from OC is $\frac{R}{\sqrt{2}}$
Extend O point to N, now consider the magnetic field due to semi infinite wire and subtract magnetic field of wire ON, this will be resultant magnetic field at point P,
$B=\frac{{\mu }_{0}i}{4\pi .\frac{R}{\sqrt{2}}}\left(cos\right(0)-cos{45}^0)=\frac{{\mu }_{0}i}{4\pi R}(\sqrt{2}-1)$