A proton with a kinetic energy of 2 . 0 eV moves into a region of uniform magnetic field of magnitude π 2 × 10 - 3 T . The angle between the direction of magnetic field and velocity of proton is 60 o . The pitch of the helical path taken by the proton is ____ cm . (Take, mass of proton = 1 . 6 × 10 - 27 kg and charge on proton = 1 . 6 × 10 - 19 C ).

Step-by-step solution

The given data is K . E = 2   eV B = π 2 × 10 - 3   T θ = 60 o The pitch of the proton is given by P = 2 π m q B × v cos θ       . . . ( i ) By using K = m v 2 2 ⇒ v = 2 K m Substituting the values in equation (i) = 2 π × 2 m K E × 1 2 × 2 1 . 6 × 10 - 19 × π × 10 - 3 = 2 × 2 × 1 . 6 × 10 - 27 × 2 × 1 . 6 × 10 - 19 × 10 3 1 . 6 × 10 - 19 = 2 × 2 × 10 – 1 = 0 .