Question

A long straight wire, carrying current $I$, is bent at its midpoint to form an angle of ${45}^{\circ }$. Induction of magnetic field at point $P$, distant $R$ 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. $\frac{(\sqrt{2}+1){\mu }_{0}I}{4\sqrt{2}\pi R}$

Answer 1

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

We know that, magnetic field due to a line wire is given by,
$B_{net}=\frac{{\mu }_{0}I(\sin {\theta }_1+\sin {\theta }_2)}{4\pi \left(d\right)}$




(i)Due to horizontal wire magnetic field induction is zero.
(ii)Magnetic field induced due to inclined wire is $B=\frac{{\mu }_{0}i}{4\pi \frac{R}{\sqrt{2}}}(1-\frac{1}{\sqrt{2}})$
(Here the term of $\sin {\theta }_2$ is negative because of the position of the point with respect to the end)

$B=\frac{(\sqrt{2}-1){\mu }_{0}I}{4\pi R}$