A particle having specific charge σ is projected in xy plane with a speed v. There exists a uniform magnetic field in z-direction having a fixed magnitude B0. The field is made to reverse its direction after every interval of 2πσB0. Calculate the maximum possible separation (in m) between two positions of the particle during its course of motion. (givenvσB0=2metre) (Except magnetic force neglect any other force including gravity throughout the motion)

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A particle having specific charge $σ$ is projected in xy plane with a speed v. There exists a uniform magnetic field in z-direction having a fixed magnitude $B_{0}$ . The field is made to reverse its direction after every interval of $\frac{2 π}{σ B_{0}}$ . Calculate the maximum possible separation (in m) between two positions of the particle during its course of motion. $(g i v e n \frac{v}{σ B_{0}} = 2 m e t r e)$ (Except magnetic force neglect any other force including gravity throughout the motion)

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answer is 8.

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$A B = \frac{4 m v}{q B_{0}} = 8 m e t r e$

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