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A particle of mass m having a charge q enters into a circular region of radius R with velocity v directed towards the centre. The strength of magnetic field is B. Find the deviation in the path of the particle

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π - {Tan}^{- 1} (\frac{mv}{BqR/2})

π - 2 {Tan}^{- 1} (\frac{mv}{qBR})

π + {Tan}^{- 1} (\frac{mv}{{BqR}^{2}})

π + 2 {Tan}^{- 1} (\frac{mv}{BqR})

answer is C.

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Detailed Solution

R = radius of curvature of circular path

$δ =$ deviation in the path

In right angled triangle $CQO$

$Tan α = \frac{r}{R}$

$\Rightarrow α = {Tan}^{- 1} (\frac{mv/BV}{R})$

$δ = π - 2 α = π - 2 {Tan}^{- 1} (\frac{mv}{BqR})$

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