Question

Find the electric flux crossing the wire frame ABCD of length l, width b, and center at a distance OP=d from an infinite line of charge with linear charge density $\lambda$. Consider that the plane of the frame is perpendicular to the line OP Fig.

• A. $\frac{2N}{\pi {\epsilon }_0}ta{n}^{-1}\left(\frac{b}{2d}\right)$
• B. $\frac{N}{\pi {\epsilon }_0}ta{n}^{-1}\left(\frac{b}{2d}\right)$
• C. $\frac{3N}{\pi {\epsilon }_0}ta{n}^{-1}\left(\frac{7b}{2d}\right)$
• D. $\frac{N}{2\pi {\epsilon }_0}ta{n}^{-1}\left(\frac{b}{2d}\right)$

Answer 1

Answer: (B)

$\frac{N}{\pi {\epsilon }_0}ta{n}^{-1}\left(\frac{b}{2d}\right)$

The flux passing through the strip of area dA is
$d\varphi =EdAcos\theta$

$=\frac{\lambda }{2\pi {\epsilon }_0\sqrt{d^2+x^2}}\left(ldx\right)\frac{d}{\sqrt{d^2+x^2}}$

$d\varphi =\frac{\lambda dl}{2\pi {\epsilon }_0}\frac{d}{\left(\sqrt{d^2+x^2}\right)}$

$\therefore \varphi =\int d\varphi =\frac{\lambda dl}{2\pi {\epsilon }_0}{\int }_{-b/2}^{b/2}\frac{d}{d^2+x^2}$

$=\frac{\lambda dl}{2\pi {\epsilon }_0}\frac{1}{d}{\left[ta{n}^{-1}\frac{x}{d}\right]}_{-b/2}^{b/2}$

$=\frac{N}{2\pi {\epsilon }_0}\times 2ta{n}^{-1}\left(\frac{b}{2d}\right)$

$\varphi =\frac{N}{2\pi {\epsilon }_0}ta{n}^{-1}\left(\frac{b}{2d}\right)$