A proton and an α -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 9 0 ∘ . Then at this time, the angle between the velocity vectors of proton and α -particle is

Question

A proton and an α -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 9 0 ∘ . Then at this time, the angle between the velocity vectors of proton and α -particle is, 6 0 ∘, 9 0 ∘, 4 5 ∘, 18 0 ∘

MARKS App · Accepted Answer

Key Idea In a circular motion, a body changes its direction by 9 0 ∘ 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 α ​ ≃ 4 m p and q α ​ = 2 q p an d T_\alpha b e t h e t im e − p er i o d o f re v o l u t i o n o f \alpha − p a r t i c l e in t h e ma g n e t i c f i e l d , im g src = " h ttp s : // c d n − q u es t i o n − p oo l . g e t ma r k s . a pp / p y q / a p e ​ am ce t / f O or n ZQ 0 R W e w 7 K gU P 7 M 4 z k 9 GN u F u Wi X 5 y Y l rs C l ​ L k A . or i g ina l . f u ll s i ze . p n g " > b r > N o w , t h er a t i oo f t im e − p er i o d o f p ro t o nan d \alpha − p a r t i c l e , b y d i v i d in g Eq . ( i ) an d ( ii ) , ​ T α ​ T p ​ ​ = 2 1 ​ ⇒ T α ​ = 2 T p ​ ​ He n ce , t h e t im e − p er i o d o f \alpha − p a r t i c l e i s d o u b l eo f t h e p ro t o n , i . e . i f p ro t o n co v ers 90^{\circ} o f an g l e f ro mi t ss t a r t in g , t h e n \alpha − p a r t i c l e w i ll co v er 45^{\circ} o f t h e an g l e . 	herefore$ Hence, the correct option is (c).

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