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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 same dimensions (length ll, width ww and thickness dd ) with carrier densities n1n_{1} and n2n_{2}, respectively. Strip 11 is placed in magnetic field B1B_{1} and strip 22 is placed in magnetic field B2B_{2}, both along positive yy-directions. Then V1V_{1} and V2V_{2} are the potential differences developed between KK and MM in strips 11 and 22, respectively. Assuming that the current II is the same for both the strips, the correct option(s) is(are)

Options

  1. AIf B1=B2{B}_{1} = {B}_{2} and n1=2n2{n}_{1} = 2{n}_{2} , then V2=2V1{V}_{2} = 2{V}_{1}
  2. BIf B1=B2{B}_{1} = {B}_{2} and n1=2n2{n}_{1} = 2{n}_{2} , then V2=V1{V}_{2} = {V}_{1}
  3. CIf B1=2B2{B}_{1} = 2{B}_{2} and n1=n2{n}_{1} = {n}_{2} , then V2=0.5V1{V}_{2} = 0.5{V}_{1}
  4. DIf B1=2B2{B}_{1} = 2{B}_{2} and n1=n2{n}_{1} = {n}_{2} , then V2=V1{V}_{2} = {V}_{1}

Correct answer

A. If B1=B2{B}_{1} = {B}_{2} and n1=2n2{n}_{1} = 2{n}_{2} , then V2=2V1{V}_{2} = 2{V}_{1}

Step-by-step solution

As done in the above question
V=BInedV = \dfrac{BI}{ned}
∴V∝Bn∴ V ∝ \dfrac{B}{n}
Or V1V2=B1n2B2n1\dfrac{{V}_{1}}{{V}_{2}} = \dfrac{{B}_{1}{n}_{2}}{{B}_{2}{n}_{1}}
∴∴ if B1=B2 and n1=2n2⇒V2=2V1{B}_{1} = {B}_{2} \text{ and } {n}_{1} = 2{n}_{2} ⇒ {V}_{2} = 2{V}_{1}
& if B1=2B2 and n1=n2⇒V2=0.5V1{B}_{1} = 2{B}_{2} \text{ and } {n}_{1} = {n}_{2} ⇒ {V}_{2} = 0.5{V}_{1}
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