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Physics-

General

Easy

Question

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

$π + 2 {T a n}^{- 1} (\frac{m v}{B q R})$
$π + {T a n}^{- 1} (\frac{m v}{B q R^{2}})$
$π - 2 {T a n}^{- 1} (\frac{m v}{q B R})$
$π - {T a n}^{- 1} (\frac{m v}{B q R / 2})$

The correct answer is: $pi minus 2 T a n to the power of negative 1 end exponent invisible function application open parentheses fraction numerator m v over denominator q B R end fraction close parentheses$

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