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In a thin rectangular metallic strip a c...

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)

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=2n_2, then V_2=0.5V_1`

D

If `B_1=2B_2 and n_1=2n_2, then V_2=V_1`

Text Solution

Verified by Experts

The correct Answer is:
A, C
(a,c): As done in the above question
`V_1=(iB)/(n_1e)(1/(d_1))` and `V_2=(iB)/(n_2e)(1/(d_2))`
In this case, `(V_1)/(V_2)=(n_2B_1)/(n_1B_2)`
Hence, (a) and (c)
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