logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
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
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...