question_answer

Find the force experienced by a semicircular rod having a charge q as shown in Fig. Radius of the wire is R, and the line of charge with linear charge density \[\lambda \] passes through its center and is perpendicular to the plane of wire.

A) \[\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}\]

B) \[\frac{\lambda q}{{{\pi }^{2}}{{\varepsilon }_{0}}R}\]

C) \[\frac{\lambda q}{4{{\pi }^{2}}{{\varepsilon }_{0}}R}\]

D) \[\frac{\lambda q}{4\pi {{\varepsilon }_{0}}R}\]

Correct Answer: B

Solution :

[b] \[{{F}_{net}}=\int_{{}}^{{}}{dq\,\,E\cos \theta }\] \[=\int\limits_{-\pi /2}^{\pi /2}{\left( \frac{q}{\pi R} \right)Rd\theta \frac{\lambda }{2\pi \varepsilon R}\cos \theta }\] \[=\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}\,\int\limits_{-\pi /2}^{\pi /2}{\cos \theta d\theta =\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}\left[ \sin \theta  \right]_{-\pi /2}^{\pi /2}}\] \[=\frac{\lambda q}{2{{\pi }^{2}}{{\varepsilon }_{0}}R}[1-(-1)]=\frac{\lambda q}{{{\pi }^{2}}{{\varepsilon }_{0}}R}\]