A neutral particle is initially at rest in a uniform magnetic field B as shown in the diagram. The particle then spontaneously decays into two fragments, one with a positive charge +q and mass 3m and the other with a negative charge -q and mass m. Neglecting the interaction between the two charged particles and assuming that the speeds are much less than speed of light, determine the time (in μs) after the decay at which
the two fragments first meet. (use the following data $q = 1 μC, B = 2 πμT, m = 10^{- 15} kg) .$ Both the charges have initial velocities in x-y plane.

Difficult

Solution

Initial momentum (p) of them will be same but in opposite direction.
$R = \frac{p}{qB} = same for both .$
Let they meet at angle $θ$ , at time t. Then $Rθ = v_{0} t$ (i)
$and R (2 π - θ) = 3 v_{0} t$ (ii)
From (i) and (ii), $t = \frac{πR}{2 v_{0}} = \frac{π}{2 v_{0}} \frac{3 {mv}_{0}}{qB}$
$= \frac{3 πm}{2 qB} = \frac{3 π \times 10^{- 15}}{2 \times 10^{- 6} \times 2 π \times 10^{- 6}}$
$= \frac{3}{4} \times 10^{- 3} s = 750 μs$

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