Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2015-Paper-2 PARAGRAPH 2 In a thin rectangular metallic s…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2015-Paper-2 Single Correct MCQ
Published on: August 13, 2026

PARAGRAPH 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 , w and d, respectively. A uniform magnetic field 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 to 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.

(i) 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 thicknesses 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)

A
If w 1 = w 2 and d 1 = 2d 2 , then V 2 = 2V 1 If B 1 = B 2 and n 1 = 2n 2 , then V 2 = 2V 1
B
If w 1 = w 2 and d 1 = 2d 2 , then V 2 = V 1 If B 1 = B 2 and n 1 = 2n 2 , then V 2 = V 1
C
If w 1 = 2w 2 and d 1 = d 2 , then V 2 = 2V 1 If B 1 = 2B 2 and n 1 = n 2 , then V 2 = 0.5V 1
D
If w 1 = 2w 2 and d 1 = d 2 , then V 2 = V 1 (ii). Consider two different metallic strips (1 and 2) of same dimensions (lengths , width w and thickness d) with carrier densities n 1 and n 2 , respectively. 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) If B 1 = 2B 2 and n 1 = n 2 , then V 2 = V 1

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Text Solution

Verified by Experts
The correct answer is:
C

(i)(a,d)I 1 = I 2

neA 1 v 1 = neA 2 v 2

d 1 w 1 v 1 = d 2 w 2 v 2

Now, potential difference developed across MK

V = Bvw

& hence correct choice is A & D

(ii). (a,c)As I 1 = I 2

n 1 w 1 d 1 v 1 = n 2 w 2 d 2 v 2

Now,

Correct options are A & C

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