A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

Question

A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes (A) 8 times (B) 4 times (C) 2 times (D) 16 times

Tardigrade Admin · Accepted Answer

As Bqv=(mv2/r) or r=(mv/Bq) According to the question, v'=2v and B' =(B/2) ∴ r'=(mv'/B' q)=(m (2v)/(B /2)q)=(4mv/Bq)=4r

Q. A charged particle is moving on circular path with velocity $v$ in a uniform magnetic field $B$ , if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

$8$ times

$4$ times

$2$ times

$16$ times

As $Bq v = \frac{m v ^{2}}{r}$ or $r = \frac{m v}{Bq}$
According to the question, $v^{'} = 2 v$ and $B^{'} = \frac{B}{2}$
$∴ r^{'} = \frac{m v ^{'}}{B ^{'} q} = \frac{m ( 2 v )}{( B /2 ) q} = \frac{4 m v}{Bq} = 4 r$

Q. A charged particle is moving on circular path with velocity v in a uniform magnetic field B, if the velocity of the charged particle is doubled and strength of magnetic field is halved, then radius becomes

8 times

4 times

2 times

16 times