Question

Two non-conducting infinitely long straight wires with uniform linear change densities ${\lambda }_1$ and ${\lambda }_2$ are arrange along x and y-axis respectively. Their point of intersection is at origin. One of the field lines due to this wires in x-y plane is a straight line passing through the origin. The angle made by this line with x-axis is

• A. $ta{n}^{-1}\left(\frac{{\lambda }_1}{{\lambda }_2}\right)$
• B. $ta{n}^{-1}\left(\frac{{\lambda }_2}{{\lambda }_1}\right)$
• C. $ta{n}^{-1}\left(\sqrt{\frac{{\lambda }_1}{{\lambda }_2}}\right)$
• D. $ta{n}^{-1}\left(\sqrt{\frac{{\lambda }_2}{{\lambda }_1}}\right)$

Answer 1

The correct option is B $ta{n}^{-1}\left(\frac{{\lambda }_2}{{\lambda }_1}\right)$

Electric Field due to wire along $x$ -axis

$\overrightarrow{E_1}=\frac{{\lambda }_1}{4\pi {\epsilon }_{0}d}\hat{ȷ}$

Electric Field due to wire along $y-$ axis

${\overrightarrow{E}}_2=\frac{{\lambda }_2}{4\pi {\epsilon }_{0}d}\hat{l}$

Then, angle made by the net $\overrightarrow{E}$ with the $x$ -axis

$$\begin{array}{cc}\varphi & ={\tan }^{-1}\left(\frac{E_2}{E_1}\right)\\ \varphi & ={\tan }^{-1}(\frac{{\lambda }_2^{\prime }}{4\pi {\epsilon }_0\alpha }\times \frac{4\pi {\epsilon }_0\alpha }{{\lambda }_1})\end{array}$$

$\varphi ={\tan }^{-1}\left(\frac{{\lambda }_2}{t_1}\right)$