Question

A long infinite current carrying wire is bent right angles as shown in figure. Find the magnetic field at point $\text{P}$ which is equidistant from both wires at $L.$

• A. $\frac{{\mu }_{0}i(\sqrt{2}+1)}{2\sqrt{2}\pi L}⨀$
• B. $\frac{{\mu }_{0}i}{2\sqrt{2}\pi L}⨂$
• C. $\frac{{\mu }_{0}i(\sqrt{2}+1)}{2\sqrt{2}\pi L}⨂$
• D. $\frac{{\mu }_{0}i}{2\sqrt{2}\pi L}⨀$

Answer 1

The correct option is C $\frac{{\mu }_{0}i(\sqrt{2}+1)}{2\sqrt{2}\pi L}⨂$

The magnetic field at a point due to a current carrying wire is,

$B=\frac{{\mu }_{0}i}{4\pi d}[\cos \alpha +\cos \beta ]$


Let, us take the given wire as $\left(1\right)$ and $\left(2\right)$ and their field at point $\text{P}$ is $B_1$ and $B_2$ respectively.


From the figure,

$B_1=\frac{{\mu }_{0}i}{4\pi d}[\cos 0^{\circ }+\cos {45}^{\circ }]$

$\therefore B_1=\frac{{\mu }_{0}i}{4\pi L}(1+\frac{1}{\sqrt{2}})⨂$

Similarly,

$B_2=\frac{{\mu }_{0}i}{4\pi d}[\cos 0^{\circ }+\cos {45}^{\circ }]$

$\therefore B_2=\frac{{\mu }_{0}i}{4\pi L}(1+\frac{1}{\sqrt{2}})⨂$

So, net magnetic field at point $\text{P}$ will be

$B_{net}=B_1+B_2=2\times \frac{{\mu }_{0}i}{4\pi L}(1+\frac{1}{\sqrt{2}})⨂$

$\therefore B_{net}=\frac{{\mu }_{0}i(\sqrt{2}+1)}{2\sqrt{2}\pi L}⨂$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{C}\right)$ is the correct answer.

Why this Question ? To understand the concept of magnetic field due to a straight conductor at a point.