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An infinitely long wire with linear char...

An infinitely long wire with linear charge density `+lambda` is bent and the two parts are inclined at angle `30^@` as shown in fig, the electric field at point P is :

A

`(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati + hatj]`

B

`(lambda)/(4 pi in_0 d) [ (1 - sqrt(3))hati + 2hatj]`

C

`(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati - 2hatj]`

D

`(lambda)/(4 pi in_0 d) [ (1 + sqrt(3))hati + 2hatj]`

Text Solution

Verified by Experts

The correct Answer is:
D
`E_1 = E_1^. = lambda/(4pi in_0 d)`
`E_1 = (lambda)/(4 pi in_0 ((sqrt(3)d)/(2))) (cos 60^@ + cos 0^@) = (sqrt(3) d)/(4 pi in_0 d)`
`:. vec(E ) = (E_1 + E_2 cos 30^@ + E_2 cos 60^@) hati + (E_1^. + E_2^1 sin 60^@ - E_2 sin 30^@) hati`

`= ((lambda)/(4pi in_0 d) + (sqrt(3)lambda)/(8pi in_0 d) + (sqrt(3) lambda)/(8 pi in_0 d)) hati + ((lambda)/(4pi in_0 d) + (sqrt(3))/(8pi in_0 d) - (sqrt(3))/(8pi in_0 d)) hatj`
`= (lambda)/(4 pi in_0 d)[(1 + sqrt(3) hati + 2hatj]`
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