A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is:

Question

A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is: (A) 1: 4 (B) 4: 1 (C) 2: 1 (D) 1: 2

Tardigrade Admin · Accepted Answer

When a charged particle enters a field, it takes a circular path. The radius of circular path is given by the relation

r = \frac{m v}{Bq}

For proton,

r_{1} = \frac{m v}{Bq}

For

α

-particle,

r_{2} = \frac{4 m \times v}{B \times 2 q} = \frac{2 m v}{Bq}

\Rightarrow r_{1} : r_{2} :: 1 : 2

Q. A proton and an alpha particle of the same enter in a uniform magnetic field which is acting perpendicular to their direction of motion. The ratio of the circular paths described by the alpha particle and proton is:

$1 : 4$

$4 : 1$

$2 : 1$

$1 : 2$

When a charged particle enters a field, it takes a circular path. The radius of circular path is given by the relation
$r = \frac{m v}{Bq}$
For proton, $r_{1} = \frac{m v}{Bq}$
For $α$ -particle, $r_{2} = \frac{4 m \times v}{B \times 2 q} = \frac{2 m v}{Bq}$
$\Rightarrow r_{1} : r_{2} :: 1 : 2$