An electron emitted by a heated cathode and accelerated through a potential difference of $2.0 \; kV$,enters a region with a 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 $30^{\circ}$ with the initial velocity.

(N/A) Given:
Magnetic field strength,$B = 0.15 \; T$
Charge on the electron,$e = 1.6 \times 10^{-19} \; C$
Mass of the electron,$m = 9.1 \times 10^{-31} \; kg$
Potential difference,$V = 2.0 \; kV = 2 \times 10^{3} \; V$
The kinetic energy of the electron is $eV = \frac{1}{2}mv^2$. Thus,the velocity $v = \sqrt{\frac{2eV}{m}}$.
$(a)$ When the magnetic field is transverse (perpendicular) to the velocity,the magnetic force provides the centripetal force: $Bev = \frac{mv^2}{r}$.
Radius $r = \frac{mv}{Be} = \frac{m}{Be} \sqrt{\frac{2eV}{m}} = \frac{1}{B} \sqrt{\frac{2mV}{e}}$.
Substituting values: $r = \frac{1}{0.15} \sqrt{\frac{2 \times 9.1 \times 10^{-31} \times 2 \times 10^3}{1.6 \times 10^{-19}}} \approx 1.01 \times 10^{-3} \; m = 1.0 \; mm$.
The trajectory is a circle of radius $1.0 \; mm$.
$(b)$ When the field makes an angle $\theta = 30^{\circ}$ with the velocity,the component of velocity perpendicular to the field is $v_{\perp} = v \sin \theta$.
The radius of the helical path is $r' = \frac{mv \sin \theta}{Be} = r \sin 30^{\circ} = 1.0 \; mm \times 0.5 = 0.5 \; mm$.
The trajectory is a helix with radius $0.5 \; mm$ and pitch $p = v \cos \theta \times T = v \cos \theta \times \frac{2\pi m}{Be}$.