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A particle of charge q and mass m is released from the origin with a velocity $v=a \hat{i}$ in a uniform magnetic field $B=b \hat{k}$ . The particle will cross the y-axis of a point whose y-coordinate is
Option: 1
$y=\frac{m a}{q b}$

Option: 2
$y=\frac{2 m a}{q^b}$

Option: 3
$y=-\frac{m a}{q^6}$

Option: 4
$y=-\frac{2 m a}{q b}$

Answers (1)

best_answer

$r=\frac{m v}{q B}=\frac{m a}{q b}$

Thus $y=-2$

$\gamma=-\frac{2 m a}{qb}$

Posted by

Deependra Verma

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