Question

The magnetic field at point $\text{P}$ due to the arrangement shown in figure is

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

Answer 1

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


The net magnetic field at point $\text{P}$ is given by,

${\overrightarrow{B}}_P={\overrightarrow{B}}_1+{\overrightarrow{B}}_2$

Where,
$B_1$ and $B_2$ are the magnetic fields due to wire - $1$ and wire - $2$ respectively.

Point $\text{P}$ is $\frac{d}{\sqrt{2}}$ distance from both wire -$1$ and wire - $2$ respectively, and field due to both wires at point $\text{P}$ is perpendicular to the plane of wires and pointing inwards.

$\therefore B_1=B_2=B$

$B_{net}=2B$

The field due to semi-infinite wire is given by,

$B=\frac{{\mu }_{0}I}{4\pi d}[\sin {\theta }_1+\sin {\theta }_2]$

$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi \frac{d}{\sqrt{2}}}[\sin \frac{\pi }{2}-\sin \frac{\pi }{4}](\because {\theta }_1={90}^{\circ },{\theta }_2=-{45}^{\circ })$

$\Rightarrow B=\frac{{\mu }_{0}I}{4\pi \frac{d}{\sqrt{2}}}[1-\frac{1}{\sqrt{2}}]$

$B=\frac{{\mu }_{0}I}{4\pi d}[\sqrt{2}-1]$

$\therefore B_{net}=2B=\frac{2{\mu }_{0}I}{4\pi d}[\sqrt{2}-1]$

$B_P=\frac{{\mu }_{0}I}{2\pi d}[\sqrt{2}-1]$

Why this Question ? Key point - The magnetic field due to a finite current carrying wire is given by, $B=\frac{{\mu }_{0}I}{4\pi d}[\sin {\theta }_1+\sin {\theta }_2]$ Caution : The angles ${\theta }_1$ and ${\theta }_2$ should be taken very carefully.