Question

Question 4.19: An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of 30º with the initial velocity.

Class 12 - Physics - Moving Charges And Magnetism Moving Charges And Magnetism Page-171

Answer 1

A electron which is accelerated by a potential difference of 2kV will have a kinetic energy gained 200 eV.
$E=\frac{1}{2}m_ev^2$
[ 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.
$r=\frac{m_ev}{qB}$
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

Answer 2

Explanation:

(a)Kinectic energy of electron = eV

also, eV=21mv2

⇒v=2eV/m−(1) 

also, BeV=mv2/r

⇒r=mV/Be

using equation, (1),r=1 mm

(b)θ=30o

v1=vsinθ

r1=mvsinθ/Be=0.5mm