1.

linear charge density of current carrying wire of infinite length is 4(muC)/m. Electric field intensity at distance 3.6cm from wire is............ [therefore 1/(4piepsilon_(0)) = 9 xx 10^(9) SI]

Answer»

`2 xx 10^(6)` V/m
`10^(6)` V/m
`10^(5)` V/m
`2 xx 10^(5)` V/m

Solution :Electric field by current carrying wire at distance V:
`E = lambda/(2piepsilon_(0)).1/r =(2lambda)/(4piepsilon_(0)r) = (2klambda)/r`
`therefore E = (2 xx 9 xx 10^(9) xx 4 xx 10^(-6))/(3.6 xx 10^(-2))`
`therefore E = 2 xx 10^(6)` V/m


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