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A conductor of length L is placed perpendicular to a horizontal uniform magnetic field B. Suddenly a certain amount of charge is passed through it, then it is found to jump to a height h. The amount of charge that passes through the conductor is

DIVYA SHARMA , 7 Years ago
Grade 12
anser 1 Answers
ROSHAN MUJEEB

Last Activity: 4 Years ago

Linera impulse =mv
or F△t=m*√2gh or (ilB)△t=μ√2gh
But i△t=△q
∴(△q)(lB)=m√2gh−−−
Hence△q=m√2gh/BI

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