JEE AdvancedPhysicsMagnetismMultiple correct+4 / −2
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 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 and appearance of equal and opposite charges on the face opposite PQRS. A potential difference along the z-direction is thus developed. Charge accumulation continues until 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 (length l, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 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 options is/are :
Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d) with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed in magnetic field B2, both along positive y-directions. Then V1 and V2 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 options is/are :- AIf B1 = B2 and n1 = 2n2, then V2 = 2V1
- BIf B1 = B2 and n1 = 2n2, then V2 = V1
- CIf B1 = 2B2 and n1 = n2, then V2 = 0.5V1
- DIf B1 = 2B2 and n1 = n2, then V2 = V1
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