A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is

Question

A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is (A) 1: 4 (B) 1:√2 (C) 1: 1 (D) 1: 2

Tardigrade Admin · Accepted Answer

For a moving charge in a perpendicular magnetic field, (m v2/r)=Bqv ⇒ r = (m v/B q) = (p/B q) = (√2 m K/q B) (rp/rd)=√(mp/md)= √(1/2)=(1/√2)

Q. A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is

$1 : 4$

$1 : 2$

$1 : 1$

$1 : 2$

For a moving charge in a perpendicular magnetic field,
$\frac{m v ^{2}}{r} = Bq v$
$\Rightarrow r = \frac{m v}{Bq} = \frac{p}{Bq} = \frac{2 m K}{qB}$
$\frac{r _{p}}{r _{d}} = \frac{m _{p}}{m _{d}} = \frac{1}{2} = \frac{1}{2}$

Q. A proton and a deuteron with the same initial kinetic energy enter a magnetic field in a direction perpendicular to the direction of the field. The ratio of the radii of the circular trajectories described by them is

1:4

1:2​

1:1

1:2

For a moving charge in a perpendicular magnetic field, rmv2​=Bqv ⇒r=Bqmv​=Bqp​=qB2mK​​ rd​rp​​=md​mp​​​=21​​=2​1​

$1 : 4$

$1 : 2$

$1 : 1$

$1 : 2$

For a moving charge in a perpendicular magnetic field,
$\frac{m v ^{2}}{r} = Bq v$
$\Rightarrow r = \frac{m v}{Bq} = \frac{p}{Bq} = \frac{2 m K}{qB}$
$\frac{r _{p}}{r _{d}} = \frac{m _{p}}{m _{d}} = \frac{1}{2} = \frac{1}{2}$