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.`

Explanation

(a) The current density j is given by
`j=i/A=50/(0.02xx1.0xx10^-2)=25xx10^5 Am^-2`
The drift velocity `v_d` is given by
`v_d=j/("ne")=(25xx10^5)/(1.1xx10^29xx1.6xx10^-19)=1.4xx10^-4ms^-1`
This velocity will be in outward direction.
(b) Magnetic force is given by `F=ev_dB`
`=(1.6xx10^-19)(1.4xx10^-4)(2)`
`=4.5xx10^-23N` (in downward direction)
(c) Force due to applied electric field should balance the magnetic force. For this we have `Ee=F`
or `E=F/e=(4.5xx10^-23)/(1.6xx10^-19)=2.8xx10^-4Vm^-1`
(in downward direction)
(d) Necessary voltage to produce this field is
`V=Eh=(2.8xx10^-4)(0.02)=5.6xx10^-6V.`
This voltage should be applied between top and bottom
sides with top voltage positive and bottom negative.
(e) The magnitude of `E_H` should be equal to E as calculated
in part (c). So, `E_H=2.8xx10^-4Vm^-1` and this will be in
downward direction.