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Finding Average Velocity

Consider two ...

Question

Consider two different metallic strips (1 and 2) of same dimensions (length $l$ , 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.

B_{1} = B_{2}

and

n_{1} = 2 n_{2}

, then

V_{2} = 2 V_{1}

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B_{1} = B_{2}

and

n_{1} = 2 n_{2}

, then

V_{2} = V_{1}

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B_{1} = 2 B_{2}

and

n_{1} = n_{2}

, then

V_{2} = 0.5 V_{1}

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B_{1} = 2 B_{2}

and

n_{1} = n_{2}

, then

V_{2} = V_{1}

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Solution

The correct options are
A If $B_{1} = B_{2}$ and $n_{1} = 2 n_{2}$ , then $V_{2} = 2 V_{1}$
C If $B_{1} = 2 B_{2}$ and $n_{1} = n_{2}$ , then $V_{2} = 0.5 V_{1}$
$V_{m} - V_{k} = \frac{I B}{n e d}$
For the case of option A
$n_{1} = 2 n_{2}, B_{1} = B_{2}$
$V_{2} = 2 V_{1}$
For the case of option C
$B_{1} = 2 B_{2}, n_{1} = n_{2}$
$V_{1} = 2 V_{2}$
Thus $V_{2} = 0.5 V$

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The correct options are A If B1=B2 and n1=2n2, then V2=2V1 C If B1=2B2 and n1=n2, then V2=0.5V1Vm−Vk=IBnedFor the case of option An1=2n2,B1=B2V2=2V1For the case of option CB1=2B2,n1=n2V1=2V2Thus V2=0.5V