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 same dimensions `n_(1) and n_(2)`, repectrively . Strip 1 is placed in magnetic field `B_(1)` and strip 2 is placed in magnetic field `B_(2)` , both along positive ` y- directions`. Then ` V_(1) and V_(2)` are the potential differences developed between `K and M` in strips 1 and 2 , respectively . Assuming that the current `I` is the same for both the strips, the correct option(s) is (are)

Options

A. If `B_(1)=B_(2)` and `n_(1)=2n_(2)`,then `V_(2)=2V_(1)`
B. If `B_(1)=B_(2)` and `n_(1)=2n_(2)`,then `V_(2)=V_(1)`
C. If `B_(1)=2B_(2)` and `n_(1)=n_(2)`,then `V_(2)=0.5V_(1)`
D. If `B_(1)=2B_(2)` and `n_(1)=n_(2)`,then `V_(2)=V_(1)`

Answer

A,C

Explanation

`V_(M)-V_(K)=(IB)/("ned")`
`(A)n_(1)=2n_(2) B_(1)=B_(2) rArrV_(2)=2V_(1)` correct
`(C) B_(1)=2B_(2) n_(1)=n_(2)`
`V_(1)=2V_(2) rArr V_(2)=0.5 V` correct