Question

Two long, thin, parallel conductors are kept very close to each other, without touching. One carries a current `i` and the other has charge `lambda`, per unit length. An electron moving parallel to the conductor is undeflected. Let `c =` velocity of right
(i) `v = (lambda c^(2))/(i)`
(ii) `v = (i)/(lambda)`
(iii) `c = (i)/(lambda)`
(iv) the electron may be at any distance from the conductor
A. (i), (ii)
B. (ii), (iii)
C. (i), (iii)
D. (i), (iv)

Answer 1

Correct Answer - D

Since electron is undeflected i.e, net force on it is zero.
The electric field at `P` due to wire
`E = (lambda)/(2pi in_(0) r)`
Electric force on electron, `F_(e) = eE = (e lambda)/(2pi in_(0) r)`, towards left
The magnetic force on electron should be towards right and for this, magnetic field due to wire should be inside the plane of paper.
`F_(m) = Bev = (mu_(0) iev)/(2pi r)`
`F_(e) = F_(m)`
`(e lambda)/(2pi in_(0) r) = (mu_(0) iev)/(2pi r) rArr v = (lambda)/(mu_(0) in_(0) i) = (lambda c^(2))/(i)`
Since `c = (1)/(sqrt(mu_(0) in_(0))`
In balancing force, `r` cancels out.