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
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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