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QJEEAdv15Ph1d7d: JEE Advanced 2015 Physics - Class 12 JEEAdv15Ph1d7d Solution: Consider two different metallic strips (1 and 2) of same…

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 :

Multiple Choice Options

  • A If B1 = B2 and n1 = 2n2, then V2 = 2V1
  • B If B1 = B2 and n1 = 2n2, then V2 = V1
  • C If B1 = 2B2 and n1 = n2, then V2 = 0.5V1
  • D If B1 = 2B2 and n1 = n2, then V2 = V1

Step-by-Step Verified Solution

 and amp;V1=i B1n1edamp; V2=i B2n2edamp;V1 V2=B1 B2n2n1\text { and } \quad \begin{aligned}& \mathrm{V}_1 =\frac{i \mathrm{~B}_1}{n_1 e d} \\\\& \mathrm{~V}_2 =\frac{i \mathrm{~B}_2}{n_2 e d} \\\\& \frac{\mathrm{V}_1}{\mathrm{~V}_2} =\frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1}\end{aligned}

For option (A) :

V1 V2=12\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{1}{2}

and B1 B2n2n1=1×12=12\frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1}=1 \times \frac{1}{2}=\frac{1}{2}

For option (B) :

V1 V2amp;=1 and B1 B2n2n1amp;=1×12=12\begin{aligned}\frac{\mathrm{V}_1}{\mathrm{~V}_2} & =1 \\\\\text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =1 \times \frac{1}{2}=\frac{1}{2}\end{aligned}

For option (C) :

V1 V2amp;=2 and B1 B2n2n1amp;=2×1=2\begin{aligned}\frac{\mathrm{V}_1}{\mathrm{~V}_2} & =2 \\\\\text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =2 \times 1=2\end{aligned}

For option (D) :

V1 V2amp;=1 and B1 B2n2n1amp;=2×1=2\begin{aligned}\frac{\mathrm{V}_1}{\mathrm{~V}_2} & =1 \\\\\text { and } \frac{\mathrm{B}_1}{\mathrm{~B}_2} \frac{n_2}{n_1} & =2 \times 1=2\end{aligned}

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