%3Cp%3EA%20small%20particle%20of%20mass%202g%20and%20charge%2010%20micro%20Coulo

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%3Cp%3EA%20small%20particle%20of%20mass%202g%20and%20charge%2010%20micro%20Couloumb%20passes%0Athrough%20the%20origin%20with%20a%20velcoity%20of%202%20m%2Fs%20in%20the%20%2B%20x-direction%0Ainto%20a%20uniform%20magnetic%20field%20parallel%20to%20the%20z-axis%20and%20a%20uniform%0Agravitational%20field%20of%201g%20in%20the%20y-direction.%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(a)%20If%20there%20is%20no%20electrical%20field%20and%20the%20particle%0A%20continues%20moving%20in%20a%20straight%20line%20at%20constant%20speed%2C%20what%0Ais%20the%20magnetic%20force%20on%20the%20particle%3F%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(b)%20What%20is%20the%20magnitude%20and%20direction%20of%20the%20magnetic%0Afield%3F%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(c)%20If%20the%20particle%20attained%20its%20speed%20by%20sliding%20down%20a%20ramp%0Abefore%20entering%20the%20magnetic%20field%2C%20what%20was%20the%20height%20of%20the%0Aramp%3F%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3EPlease%20explain%20thouroughly.%20Especially%20the%20part%20about%20the%0Agravitational%20field%2C%20is%20that%20included%20in%20calculations%3F%20or%20just%20put%0Athere%20to%20throw%20us%20off%3F%26nbsp%3B%3C%2Fp%3E%0A%3Cp%3EThanks%20in%20Advance!!%3C%2Fp%3E%0A

Explanation / Answer

Part A)

The magnetic force must balance the weight

F = mg = (.002)(9.8)

F = .0196 N

Part B)
F = qvB

.0196 = (10 X 10^-6)(2)(B)

B = 980 Tesla

Part C)

PE = KE

mgh = .5mv^2

(9.8)(h) = (.5)(2)^2

h = .204 m

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