A particle of charge +q and mass m is in a uniform magnetic field $B = B_{0} \hat{z} .$ The initial position of the particle is $x = 0, y = 0, z = 0$ , the magnitude of the momentum of the particle is $p_{0}$ . The particle is so projected that it crosses the point $[x = \frac{p_{0}}{q B_{0}}, y = 0, z = 0]$ . What is the time (in sec) after which it reaches this point ? Take $\frac{m}{q} = \frac{12}{π} S I$ units and $B = 0.2 T .$

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answer is 20.

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Detailed Solution

$θ = ω t = \frac{q B t}{m} \sin \frac{θ}{2} = \frac{a}{2 R} \Rightarrow \frac{θ}{2} = \sin^{- 1} (\frac{a q B}{2 p_{0}}) t = \frac{2 m}{q B} \sin^{- 1} (\frac{a q B}{2 p_{0}}) = \frac{2 m}{q B} \times \frac{π}{6} = \frac{12}{π \times 1 / 5} \times \frac{π}{6} = 20 \sec$