Question

an electron and a proton have the same kinetic energy upon entering a region of constant magnetic field and their velocity vectors are perpendicular to the magnetic field. Suppose the magnetic field is strong enough to allow the particles to circle in the field. What is the ratio of the radii of their circular paths rp/rc?

Answer 1

`= 2.3 *10 ^(-2)`

Step-by-step explanation:

using formula for radius

r =
`(mv)/(qB)`

mv = qBr (momentum)......(i)

where m is mass v is velocity

q is charge B is magnetic field

using relation b/w kinetic energy and momentum

`p = √( 2 KE m)` ............(ii)

where p is momentum and

K.E is kinetic energy

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

from (i) and (ii) equation we get

`qBr = √(2 KE m)`

`r = (√(2KE m) )/(qB)`

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

so we get

r will be directly proportional to square root of mass m

`(r_e)/(r_p) = \sqrt{(m_(e))/(m_p) }`

`= \sqrt{\frac{9.1 * 10^(-31)}{1.67 *10^{{-27}}}`

`= 2.3 *10 ^(-2)`