Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Electric field intensity due to a semi- infinite wire at a point A as shown is :

$(A)\;\frac{\lambda}{4 \pi \in _0 a} \sin \frac{\pi}{4} \hat i +\frac{\lambda}{4 \pi \in _0 a} \cos \frac{\pi}{4} \hat j \\ (B)\;\frac{\lambda}{4 \pi \in _0 a} (1+\sin \frac{\pi}{4}) \hat i +\frac{\lambda}{4 \pi \in _0 a} \cos \frac{\pi}{4} \hat j \\ (C)\; \frac{\lambda}{4 \pi \in _0 a} (1-\sin \frac{\pi}{4}) \hat i +\frac{\lambda}{4 \pi \in _0 a} \cos \frac{\pi}{4} \hat j \\ (D)\;\frac{\lambda}{4 \pi \in _0 a} \sin \frac{\pi}{4} \hat i +\frac{\lambda}{4 \pi \in _0 a} \cos \frac{\pi}{4} \hat j $

Can you answer this question?

1 Answer

0 votes
$\overrightarrow{E}$ due to straight wire
$E_x= \large\frac{\lambda}{4 \pi \in _0 a} $$(\sin \theta_2+ \sin \theta_1)$
$E_y= \large\frac{\lambda}{4 \pi \in _0 a} $$(\cos \theta_1- \cos \theta_2)$
$\theta_1=\large\frac{-\ pi}{4}$
$\theta_2=\large\frac{\ pi}{2}$
So,$\large \frac{\lambda}{4 \pi \in _0 a} $$(1-\sin \frac{\pi}{4}) \hat i +\large\frac{\lambda}{4 \pi \in _0 a} $$\cos \frac{\pi}{4} \hat j$
Hence C is the correct answer.
answered Jan 13, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App