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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2015 · Paper 2 · Physics · Magnetic Effects of Current

Paragraph: In a thin rectangular metallic strip a constant current...

Paragraph: In a thin rectangular metallic strip a constant current II flows along the positive xx-direction, as shown in the figure. The length, width and thickness of the strip are l,wl, w and dd, respectively. A uniform magnetic field B⃗\vec{B} is applied on the strip along the positive yy-direction. Due to this, the charge carriers experience a net deflection along the zz-direction. This results in accumulation of charge carriers on the surface PQRSP Q R S and appearance of equal and opposite charges on the face opposite to PQRSP Q R S. A potential difference along the zz-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. Paragraph: In a thin rectangular metallic strip a constant current I flows along the positive x -direction, as shown in
Paragraph: In a thin rectangular metallic strip a constant current I flows along the positive x -direction, as shown in

Question: Consider two different metallic strips (11 and 22) of the same material. Their lengths are the same, widths are w1w_{1} and w2w_{2} and thicknesses are d1d_{1} and d2d_{2}, respectively. Two points KK and MM are symmetrically located on the opposite faces parallel to the x−yx-y plane (see figure). V1V_{1} and V2V_{2} are the potential differences between KK and MM in strips 11 and 22, respectively. Then, for a given current II flowing through them in a given magnetic field strength BB, the correct statement(s) is(are)

Options

  1. AIf w1=w2{w}_{1} = {w}_{2} and d1=2d2{d}_{1} = 2{d}_{2} , then V2=2V1{V}_{2} = 2{V}_{1}
  2. BIf w1=w2{w}_{1} = {w}_{2} and d1=2d2{d}_{1} = 2{d}_{2} , then V2=V1{V}_{2} = {V}_{1}
  3. CIf w1=2w2{w}_{1} = 2{w}_{2} and d1=d2{d}_{1} = {d}_{2} , then V2=2V1{V}_{2} = 2{V}_{1}
  4. DIf w1=2w2{w}_{1} = 2{w}_{2} and d1=d2{d}_{1} = {d}_{2} , then V2=V1{V}_{2} = {V}_{1}

Correct answer

A. If w1=w2{w}_{1} = {w}_{2} and d1=2d2{d}_{1} = 2{d}_{2} , then V2=2V1{V}_{2} = 2{V}_{1}

Step-by-step solution

qvdB=qE=qVwq{v}_{d}B = qE = \dfrac{qV}{w}
⇒V=wB×vd=wBInewd=BIned⇒ V = wB \times {v}_{d} = \dfrac{wBI}{newd} = \dfrac{BI}{ned}
∴V1V2=d2d1∴ \dfrac{{V}_{1}}{{V}_{2}} = \dfrac{{d}_{2}}{{d}_{1}}
∴∴ if d1=2d2⇒V2=2V1{d}_{1} = 2{d}_{2} ⇒ {V}_{2} = 2{V}_{1}
& if d1=d2⇒V1=V2{d}_{1} = {d}_{2} ⇒ {V}_{1} = {V}_{2}
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