An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by me and mp respectively, then the ratio of the radii (electron to proton) of their circular orbits is

Question

An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by me and mp respectively, then the ratio of the radii (electron to proton) of their circular orbits is (A) √(mp/me) (B) √(me/mp) (C) (me/mp) (D)  1

Tardigrade Admin · Accepted Answer

Radius of circular path in a uniform magnetic field

r = \frac{m v}{qB}

Kinetic energy of the charged particle

K = \frac{1}{2} m v^{2}

v^{2} = \frac{2 K}{m}

\Rightarrow v = \frac{2 K}{m}

∴ r = \frac{m}{qB} \frac{2 K}{m}

r = \frac{2 K m}{qB}

For the same value of

K

and

B

r \propto \frac{m}{q}

∴ \frac{r _{e}}{r _{p}} = \frac{m _{e}}{m _{p}} \times \frac{q _{p}}{q _{e}}

\frac{r _{e}}{r _{p}} = \frac{m _{e}}{m _{p}} \times \frac{e}{e} = \frac{m _{e}}{m _{p}}

Q. An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by $m_{e}$ and $m_{p}$ respectively, then the ratio of the radii (electron to proton) of their circular orbits is

$\frac{m _{p}}{m _{e}}$

$\frac{m _{e}}{m _{p}}$

$\frac{m _{e}}{m _{p}}$

$1$

Q. An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by me​ and mp​ respectively, then the ratio of the radii (electron to proton) of their circular orbits is

me​mp​​​

mp​me​​​

mp​me​​

1

Q. An electron and a proton, both having the same kinetic energy, enter a region of uniform magnetic field, in a plane perpendicular to the field. If their masses are denoted by $m_{e}$ and $m_{p}$ respectively, then the ratio of the radii (electron to proton) of their circular orbits is

$\frac{m _{p}}{m _{e}}$

$\frac{m _{e}}{m _{p}}$

$\frac{m _{e}}{m _{p}}$

$1$