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An electron with a velocity of 10^7 m/s enter into a region of magnetic flux density of 0.10T,the angle between the direction of the field and the initial path of electron being 25°.Find the axial distance between two turn of the helical path​

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Answer:

F = q v B sin Θ       describes force on the electron

F = m v^2 / R          describes force required to keep electron in circular path

q v B sin Θ = m v^2 / R

R = m v / (q B sin Θ)

R = 9.1E-31 * 1E7 / (.1 * 1.6E-19 * .423)

R = 9.1 / 6.77 * E-3 = .00134 m = .134 cm

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