82. (a) We are given 5ˆˆi(610T)ixBB−==×G, so that GGvB vByx×=−±k where vy= 4×104m/s. We note that the magnetic force on the electron is −−evBbgej±k and therefore points in the +±k direction, at the instant the electron enters the field-filled region. In these terms, Eq. 28-16 becomes rmveBeyx0 0038.m.(b) One revolution takes T= 2πr/vy= 0.60 µs, and during that time the “drift” of the electron in the xdirection (which is the pitchof the helix) is ∆x = vxT= 0.019 m where
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