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 = 1µC, B = 2πµT, m = 10–15 kg). Both the charges have initial velocities in x-y plane.

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Question

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 = 1µC, B = 2πµT, m = 10^–15 kg). Both the charges have initial velocities in x-y plane.

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$r = \frac{p}{qB} = same,$ $T_{+} = \frac{2 {πm}_{+}}{qB} = \frac{6 πm}{qB},$ $T_{-} = \frac{2 πm}{qB}$
as T₊ = 3T_–, They will meet at θ =π/2

q = 1 µC, B = 2πµT , m = 10^–15 kg

The time is $= \frac{T_{+}}{4} = \frac{6 πm}{4 qB} = \frac{6 \times π \times 10^{- 15}}{4 \times 1 \times 10^{- 6} \times 2 π \times 10^{- 6}}$
= 0.75 × 10^–3 S = 750 µS