A current i, indicated by the crosses in figure, is established in a strip of copper of height h and width w. A uniform field of magnetic induction `B` is applied at right angle to the strip.
(a) Calculate the drift velocity `v_d` of the electrons.
(b) What are the magnitude and direction of the magnetic force F acting on the electrons?
(c) What should the magnitude and direction of a homogeneous electric field E be in order to counterbalance the effect of mangetic field?
(d) Calculate voltage V necessary between two sides of the conductor in order to creat this field `E` Between which sides of the conductor would this voltage have to be applied?
(e) If no electric field is applied from the outside, the electrons will be pushed somewhat to one side and therefore will give rise to a uniform electric field `E_H` across the conductor until the forces of this electrostatic field `E_H` balance the magnetic forces encountered in part (b). What will be the magnitude and direction of field `E_H`?
Assume that n, the number of conductor electrons per unit volume is `1.1xx10^(29) m^-3, h = 0.02 m, w= 0.1 cm, i = 50 A, and B=2 T.`

Answer

(a) `1.4 xx 10^(-4) m//s` (b) `4.5 xx 10^(-23) N` (down) (c ) `2.8 xx 10^(-4) V//m (down) (d) `5.7 xx 10^(-6) V ("top" + "bottom"-) (e) same as (c ) .

Explanation

`I = "ne" AV darr implies V_(d) = (I)/(n eA)`
`V_(d) = (50)/(1.1 xx 10^(29) xx 1.6 xx 10^(-19) xx 2 xx 10^(5)) = 1.4 xx 10^(-4) m//s`
(b) `oversetrarr(F_(e)) = e(oversetrarrV_(d) xx oversetrarrB)implies F_(e) = e VdarrB`
` = 1.6 xx 10^(-19) xx 1.4 xx 10^(-4) xx 2N = 5 xx 10^(-23) N darr`
(c) `F_(e) =F_(B)`
`eE = eV_(d)B`
`E = 2.8 xx 10^(-4) V//mdarr`
(d) `V =E xx h`
` = 2.8 xx 10^(-4) xx 0.2`
` =5.6 xx 10^(-6) "volt"`
(e) same as (c)
.