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A square loop of side 'l' having uniform...

A square loop of side 'l' having uniform linear charge density '`lambda`' is placed in 'xy' plane as shown in the figure. There is a non uniform electric field `vec E= a/l (x+l) hat i` where a is a constant. Then resultant electric force on the loop if `l=10cm, a=2N//C` and charge density `lambda = 2muC//m` is:

A

`4xx10^(-6) N`

B

`4xx10^(-5)N`

C

`3xx10^(-6) N`

D

`3xx10^(-5)N`

Text Solution

Verified by Experts

The correct Answer is:
A

E at AB=`a/l (l+l)=2a :. ` F on AB `=2 a lambda l`
E at CD `=a/l(2l+l) =3a :. F "on" CD=3 a lambdal`
On BC and AD electric field is uniform `:'` x is not constant. But on BC and AD electric field will have the same type of variation.
`F_(AD)=F_(BC)=int_(x=l)^(2l)(lambda dx).a/l (x+l)`
`=(alambda)/l[(x^(2))/2+lx]_(l)^(2l)=(alambda)/l[(3l^(2))/2+l^(2)]=5/2 a lambdal`
`:.` Total force on the loop =`2a lambdal+3a lambdal+2(5/2 a lambdal)`
`F=10 a l lambda`
Using values `F=4xx10^(-5) N`
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