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

`(mv)/(qB)`Answer:

a) ratio of radii for same momentum = 1

b) ratio of radii for same energy = 2.3 ×
`10^(-2)`

Step-by-step explanation:

using formula for radius

r =
`(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 ∝
`(1)/(qB)`

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

`(r_(1) )/(r_(2) ) =(1)/(1)`

b) using relation b/w kinetic energy and momentum

p =
`√(2 KEm)` ............(1)

where p is momentum and

K.E is kinetic energy

we know
`r =(mv)/(qB)`

`mv =qBr`

`p = mv` =
`qBr`

from first equation

`qBr = √(2KEm)`

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

so we get

`(r_(e) )/( r_(p) )`
`= \sqrt{(m_(e) )/(m_(p ) )`

`(r_(e) )/(r_(p) )`
`= \sqrt{(9.1 x 10^(-31) )/(1.67 x 10^(-27) ) } = 2.3 x 10 ^(-2)`