Question

An electron enters a uniform magnetic field B=0.23x10-2 Wb/m2 at 450 angle to B. Determine the radius and pitch of helical path. Assume electron speed to be 3x107 m/s.

Answer 1

Given:- B=0.23x10^-2 Wb/m²

v=3x10⁷ m/s

e=1.6x10^-19 C

m=9.1x10^-31 kg

Formula: R=mv/eB

`R=(9.1×10^(−31)×3×10^7)/(1.6×10^(−19)×0.23×10^(−2))`

R=0.074 m

Thus the radius of helical path is 0.074m.