Electron accelerated through a potential difference of 2kV enter a region of uniform magnetic field of 0.1 T. Find radius.
Answers
Answer: A electron which is accelerated by a potential difference of 2kV will have a kinetic energy gained 200 eV.
[ here, me is mass of electron and v is the velocity of electron]
200eV = 1/2 × 9.1 × 10^-31 × v²
200 × 1.6 × 10^-19 = 1/2 × 9.1 × 10^-31 × v²
after solving it , we get v = 2.66 × 10^7 m/s
(a) when the electron enters in the uniform magnetic field which is normal to the velocity of electron follows a circular path radius.
here, me = 9.1 × 10^-31 kg ,v = 2.66 × 10^7 m/s,
q = 1.6 × 10^-19 and B = 0.15T
so, r = (9.1 × 10^-31 × 2.66 × 10^7)/(1.6 × 10^-19 × 0.15)
r = 99.75 × 10^-5 m ≈ 1mm
(b) when the magnetic field makes an angle 30° with the initial velocity the trajectory of the electron becomes helical.
e.g., r = mvsinA/qB
A is the angle between magnetic field and initial velocity.
or, r = (mv/qB)sinA
r = 99.75 × 10^-5 × sin30° ≈ 0.5mm
and v = vcosA = 2.66 × 10^7 × cos30°
= 2.3 × 10^7 m/s
pitch of helical path = T × vcosA
where T is time period e.g.,T= 2πm/qB
so,pitch = 2πmvcosA/qB
= (2π × 9.1 × 10^-31 × 2.3 × 10^7)/(1.6 × 10^-19 × 0.15)
= 542.5 × 10^-5 m ≈ 5.42mm
Explanation: Hope it helps you Pls mark me as brainliest