An electron is projected with velocity `v_(0)` in a unifrom electric field E perpendicular to the field . Again it is projected with velocity `v_(0)` perpendicular to a unifrom magnetic field B . If `r_(1)` is initial radius of curvature just after entering in the electric field and `r_(2)` is initial radius of curavaature just after entering in mag netic field then the raio `r_(1)//r_(2)` is equal to

Options

A. `(Bv_(0)^(2))/E`
B. `B/E`
C. `(Ev_(0))/B`
D. `(BV_(0))/E`

Answer

D

Explanation

From radius of curvature concept `((v_(0)^(2))/(R_(1)))` is perpendicular to velocity vector.

`implies (qE)/m=(v_(0)^(2))/(r_(1))impliesr_(1)=(mv_(0)^(2))/(qE)` `(r_(1))/(r_(2))=(Bv_(0))/E`