First slide

Force on a charged particle moving in a magnetic field

Question

An $α - particle$ and a proton are both simultaneously projected in opposite directions into a region of constant magnetic field perpendicular to the direction of the field. After some time it is found that the velocity of the $α - particle$ has changed in direction by 45°. Then at this time, the angle between velocity vectors of $α - particle$ and proton is

Moderate

A

90°
B

45°
C

135°
D

none

Solution

$ω = qB / m, ω_{P} = eB / m, ω_{α} = \frac{2 eB}{4 m} = ω_{P} / 2$
So proton will cover double angle in same time as that of $α - particle$ Clearly from figure angle between $v_{α} and v_{P} is 135^{0} .$

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