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There is a uniform magnetic field B normal to the xy plane. A conductor ABC has length $AB = l_{1}$ , parallel to the x-axis, and length $BC = l_{2}$ , parallel to the y-axis. ABC moves in the xy plane with velocity $v_{x} \hat{i} + v_{y} \hat{j}$ . The potential difference between A and C is proportional to

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v_{x} l_{1} - v_{y} l_{2}

v_{x} l_{1} + v_{y} l_{2}

v_{x} l_{2} + v_{y} l_{1}

v_{x} l_{2} - v_{y} l_{1}

answer is C.

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Detailed Solution

\bar{l} = l_{1} \hat{i} + l_{2} \hat{j}

$\bar{V} = V_{x} \hat{i} + V_{y} \hat{j} and \bar{B} = B \hat{k}$

$\bar{E} = \bar{l} . (\bar{V} \times \bar{B}) = (l_{1} \hat{i} + l_{2} \hat{j}) [(V_{x} \hat{i} + V_{y} \hat{j}) \times B \hat{k}]$

$= B [V_{x} l_{2} - V_{y} l_{1}]$

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