In a thin rectangular metallic strip a constant current `I` flows along the positive `x-direction` , as shown in the figure. The length , width and thickness of the strip are `l, w and d`, respectively.
A uniform magnetic field `vec(B)` is applied on the strip along the positive `y- direction` . Due to this, the charge carriers experience a net deflection along the `z- direction` . This results in accumulation of charge carriers on the surface `PQRS` ansd apperance of equal and opposite charges on the face opposite to `PQRS`. A potential difference along the `z-direction` is thus developed. Charge accumulation contiues untill the magnetic force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross- section of the strip and carried by electrons.
Consider two different metallic strips `(1 and 2)` of the same material . Their lengths are the same,widths are `w_(1) and w_(2)` and thickness are `d_(1) and d_(2)` respectively. Two points `K and M` are symmetrically located on the opposite faces parallel to the ` x-y` plane ( see figure) . `V_(1) and V_(2)` are the potential differences between `K and M` in strips ` 1 and 2`, respectively . Then, for a given current `I` flowing through them in a given magnetic field strength `B`, the correct statement(s) is (are)

Options

A. If `w_(1)=w_(2)` and `d_(1)=2d_(2)`,then `V_(2)=2V_(1)`
B. If `w_(1)=w_(2)` and `d_(1)=2d_(2)`,then `V_(2)=V_(1)`
C. If `w_(1)=2w_(2)` and `d_(1)=d_(2)`,then `V_(2)=2V_(1)`
D. If `w_(1)=2w_(2)` and `d_(1)=d_(2)`,then `V_(2)=V_(1)`

Answer

A,D

Explanation

`qvB=(q(V_(m)-V_(k)))/w ` velocity of electrons
`V_(m)-V_(k)=wvB.`
`I=neAv=ne(wd)v`
`wv=I/(ned)`
`V_(m)-V_(k)=I/(ned)B`
`(A)w_(1)=w_(2) d_(1)=2d_(2) rArr V_(2)=2V_(1)`
`(D)w_(1)=2w_(2) d_(1)=d_(2) rArr V_(1)=V_(2)`