A metallic block carrying current $I$ is subjected to a uniform magnetic induction $\overrightarrow B $ as shown in the figure. The moving charges experience a force $\overrightarrow F $ given by ........... which results in the lowering of the potential of the face ........ Assume the speed of the carriers to be $v$
IIT 1996, Diffcult
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(a) As the block is of metal, the charge carriers are electrons, so for current along positive $x-$axis, the electrons are moving along negative $x$-axis, i.e. $\overrightarrow {v\,} = - \,v\hat i$
and as the magnetic field is along the $y$-axis, i.e. $\overrightarrow B = B\hat j$
so $\overrightarrow F = q(\overrightarrow {v\,} \times \overrightarrow B )$ for this case yield $\overrightarrow F = ( - \,e)[ - v\hat i \times B\hat j)]$
i.e., $\overrightarrow F = evB\hat k$ [As $\hat i \times \hat j = \hat k$]
As force on electrons is towards the face $ABCD$, the electrons will accumulate on it an hence it will acquire lower potential.
and as the magnetic field is along the $y$-axis, i.e. $\overrightarrow B = B\hat j$
so $\overrightarrow F = q(\overrightarrow {v\,} \times \overrightarrow B )$ for this case yield $\overrightarrow F = ( - \,e)[ - v\hat i \times B\hat j)]$
i.e., $\overrightarrow F = evB\hat k$ [As $\hat i \times \hat j = \hat k$]
As force on electrons is towards the face $ABCD$, the electrons will accumulate on it an hence it will acquire lower potential.
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