An electron accelerated by a potnetial difference `V = 1.0 kV` moves in a unifrom magentic field at angle `alpha = 30^(@)` to the vector `B` whose modulus is `B = 29 mT`. Find the pithch of the helical trajectroy of the electron.

An electron accelerated by a potnetial difference `V = 1.0 kV` moves i

1.	An electron accelerated by a potnetial difference `V = 1.0 kV` moves in a unifrom magentic field at angle `alpha = 30^(@)` to the vector `B` whose modulus is `B = 29 mT`. Find the pithch of the helical trajectroy of the electron.
Answer» `T = eV = (1)/(2) mv^(2)` (The given potential difference is not large enough to cause significant deviations from the nonletivistic formula). Thus,. `v = sqrt((2eV)/(m))` So, `v_(\|\|) = sqrt((2eV)/(m)) cos alpha, v_(_\|_) = sqrt((2eV)/(m)) sin alpha` Now, `(m v_(_\|_)^(2))/(r) = B ev_(_\|_)` or, `r = (m v_(_\|_))/(Be)`, and `T = (2pi)/(v_(_\|_)) = (2pi m)/(Be)` Pitch `p = v_(\|\|) T = (2pi m)/(Be) sqrt((2eV)/(m)) cos alpha = 2pi sqrt((2mV)/(eB^(2))) cos alpha`

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An electron accelerated by a potnetial difference `V = 1.0 kV` moves in a unifrom magentic field at angle `alpha = 30^(@)` to the vector `B` whose modulus is `B = 29 mT`. Find the pithch of the helical trajectroy of the electron.

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