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A Positive charge Particle of mass m and charge q is projected with velocity v as shown in fig. if radius of curvature of charge Particle in magnetic field is $R \quad(2 d<R<3 d)$ , then time lapses by charge particle in magnetic field regions is

Option: 1
$\frac{m}{2 q B}$

Option: 2
$\frac{m}{q B} \sin ^{-1}\left(\frac{2 d}{R}\right)$

Option: 3
$\frac{m}{qB}$

Option: 4
$\frac{m}{qB} \sin ^{-1}\left(\frac{d}{R}\right)$

Answers (1)

best_answer

$\sin \theta =\frac{2 d}{R} \\$
$t =\frac{\theta R}{V}$

$=\frac{m}{q B} \sin ^{-1} (\frac{2 d}{R})$

$t=\frac{m}{q B} \sin ^{-1} (\frac{2 d}{R})$

Posted by

Ritika Kankaria

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