An electron moving with kinetic energy 6.6 x 10-14 J enters a magnetic field 4 X10-3 T at right angle to it, The radius of circular path will be nearest to

evB=mv2/R or ebR=mvSquaring on both sides, we gete2B2R2=m2v2=2m12mv2=2m× K.E. ∴ R=(2m×K.E.)1/2eB=2×9⋅1×10−31×6⋅6×10−141/21.6×10−19×4×10−3= 54 cm, which is nearest to 50 cm.