A magnetic field is given. In it a proton, a deutron and an alpha particle are entering with same kinetic energy. They move along circular paths. Find radius ratio.

Open in App

Solution

We know that
$m v^{2} / r = q v B$
The radius of the cirxular path is given as
$r = m v / q B$

now as $E = (1 / 2) m v^{2}$
we have, $m v = (2 m E)^{1 / 2}$

so,the radius will be
$r = \sqrt{} (2 m E) / q B$
or
the radius of proton
$r_{p} = \sqrt{} (2 m_{d} E) / q_{p} B$
the radius of a deuteron
$r_{d} = \sqrt{} (2 m_{d} E) / q_{d} B$
the radius of an alpha particle
$r_{a} = \sqrt{} (2 m_{a} E) / q_{a} B$

we also know that
$m_{p} = 2 m_{d} = 4 m_{a}$
and
$q_{p} = q_{d} = 2 q_{a}$

so,the ratios would be
$r_{p} : r_{d} r_{a} = 1 : \sqrt{} 2 : 1$

Suggest Corrections

Similar questions

Q. A proton, a deutron and an

α

-particle with same kinetic energy enter perpendicularly in a uniform magnetic field, then the ratio of radii of their circular paths is

Q. A proton, a deutron and an

α

particle of same kinetic energy enter in a uniform magnetic field perpendicularly. The ratio of their circular path will be

Q. A proton, a deutron and an

α -

particle are accelerated through the same potential difference and then they enter a uniform normal magnetic field. If the radius of circular path of proton is

8 c m

then the radius of circular path of deutron will be

Q. A proton, deutron and an

α

-particle enter a magnetic field perpendicular to field with same velocity. What is the ratio of the radii of circular paths?

A proton and an alpha particle enter the same magnetic field which is perpendicular to their velocity. If they have same kinetic energy then ratio of radii of their circular path is

Join BYJU'S Learning Program