An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV,

Solution:

Magnetic field strength, B = 0.15 T

Charge on the electron, e = 1.6 × 10−19 C

Mass of the electron, m = 9.1 × 10−31 kg

Potential difference, V = 2.0 kV = 2 × 103 V

Thus, kinetic energy of the electron = eV

$\Rightarrow e V=\frac{1}{2} m v^{2}$

$v=\sqrt{\frac{2 e V}{m}}$   ...(i)

Where,

v = velocity of the electron

(a) Magnetic force on the electron provides the required centripetal force of the electron. Hence, the electron traces a circular path of radius r.

Magnetic force on the electron is given by the relation,

B ev

Centripetal force $=\frac{m v^{2}}{r}$

$\therefore B e v=\frac{m v^{2}}{r}$

$r=\frac{m v}{B e}$   ...(i)

From equations (1) and (2), we get

$r=\frac{m}{B e}\left[\frac{2 e V}{m}\right]^{\frac{1}{2}}$

$=\frac{9.1 \times 10^{-31}}{0.15 \times 1.6 \times 10^{-19}} \times\left(\frac{2 \times 1.6 \times 10^{-19} \times 2 \times 10^{3}}{9.1 \times 10^{-31}}\right)^{\frac{1}{2}}$

$=100.55 \times 10^{-5}$

$=1.01 \times 10^{-3} \mathrm{~m}$

$=1 \mathrm{~mm}$

Hence, the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

(b) When the field makes an angle θ of 30° with initial velocity, the initial velocity will be,

$v_{1}=v \sin \theta$

From equation (2), we can write the expression for new radius as:

$r_{1 .}=\frac{m v_{1}}{B e}$

$=\frac{9.1 \times 10^{-31}}{0.15 \times 1.6 \times 10^{-19}} \times\left[\frac{2 \times 1.6 \times 10^{-19} \times 2 \times 10^{3}}{9 \times 10^{-31}}\right]^{\frac{1}{2}} \times \sin 30^{\circ}$

$=0.5 \times 10^{-3} \mathrm{~m}$

$=0.5 \mathrm{~mm}$

Hence, the electron has a helical trajectory of radius $0.5 \mathrm{~mm}$ along the magnetic field direction.