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A particle of charge per unit mass `alpha` is released from origin with a velocity `vecv=v_(0)hati` uniform magnetic field `vecB=-B_(0)hatk`. If the particle passes through `(0,y,0)`, then `y` is equal toA. `-(2v_(0))/(B_(0)alpha)`B. `(v_(0))/(B_(0)alpha)`C. `(2v_(0))/(B_(0)alpha)`D. `-(v_(0))/(B_(0)alpha)` |
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Answer» Correct Answer - C Directin of force along `+y` direction so partical passes throught the `+y` axis at a `y = 2r = (2mv)/(rhoB) = (2 v_(0))/(B_(0) alpha)` |
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