A conductor of length $l$ and mass $m$ is placed along the east-west line on a table. Suddenly a certain amount of charge is passed through it and it is found to jump to a height $h$. The earth’s magnetic induction is $B$. The charge passed through the conductor is:
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The magnetic force is
$F=l / B$
or $\mathrm{F}=\frac{\mathrm{d} \mathrm{q}}{\mathrm{dt}} / \mathrm{B}$
or $\quad \mathrm{d} \mathrm{q}=\frac{\mathrm{Fdt}}{l \mathrm{B}}$
or $\quad \Delta q=\frac{F d t}{l B}=\frac{m v_{0}}{l B}$
But $0^{2}=v_{0}^{2}-2 g h$
$\therefore v_{0}=\sqrt{2 \mathrm{gh}}$
$\Delta q=\frac{m \sqrt{2 g h}}{l B}$
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