A proton and an -particle are simultaneously projected in opposite direction into a region of uniform magnetic field of 2 mT perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by 90^ . Then at this time, the angle between the velocity vectors of proton and -particle is

Step-by-step solution

Key Idea In a circular motion, a body changes its direction by 90^ in one-fourth of its time period. Given, magnetic field, B=2 mT Let T_p be the time-period of revolution of proton in m_ 4 m p and q_ =2 q^p and T_ be the time-period of revolution of -particle in the magnetic field, Now, the ratio of time-period of proton and -particle, by dividing Eq. (i) and (ii), aligned & T_p T_ = 1 2 & T_ =2 T_p & aligned Hence, the time-period of -particle is double of the proton, i.e. if proton covers 90^ of angle from its s