Question

An electron and a proton move in circular orbits in a plane perpendicular to a uniform magnetic field B > . Find the ratio of the radii of their circular orbits when the electron and the proton have (a) the same momentum and (b) the same kinetic energy.

Answer 1

$\frac{mv}{qB}$Answer:  

a)  ratio of radii for same momentum  = 1

b)   ratio of radii for same energy           = 2.3 × $10^{-2}$

Explanation:

using formula  for radius

                            r =   $\frac{mv}{qB}$

                       where m is mass

                        v is velocity

                        q is charge

                        B is magnetic field

a) for same momentum radius will be inversely proportional to                                              roduct of charge and magnetic and magnetic field

                       r ∝  $\frac{1}{qB}$

  since for proton and electron s charge and magnetic field is same so

                     $\frac{r_{1} }{r_{2} } =\frac{1}{1}$

b) using relation b/w kinetic energy and momentum

                 p = $\sqrt{2 KEm}$   ............(1)

where p is momentum and

   K.E is kinetic energy

 we know   $r =\frac{mv}{qB}$

                  $mv =qBr$

                  $p = mv$ = $qBr$

 from first equation  

                 $qBr = \sqrt{2KEm}$

since qB is same for electron and proton and energy is also         same(given)  

so we get

          $\frac{r_{e} }{ r_{p} }$$= \sqrt{\frac{m_{e} }{m_{p } }$

         

           $\frac{r_{e} }{r_{p} }$$= \sqrt{\frac{9.1 x 10^{-31} }{1.67 x 10^{-27} } } = 2.3 x 10 ^{-2}$