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, $V =2.0\, kV =2 \times 10^{3} \,V$

Thus, kinetic energy of the electron $=e V$

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

$v=\sqrt{\frac{2 e V}{m}}\dots(i)$

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,$ Bev$

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

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

$r=\frac{m v}{B e}\dots(ii)$

From equations $(i)$ and $(ii)$, we get

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

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

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

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

$=1\, m,m$

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

$(b)$ When the field makes an angle $\theta$ of $30^{\circ}$ with initial velocity, the initial velocity will be, $v_{1}=v \sin \theta$

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

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

$=\frac{m v \sin \theta}{B e}$

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

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

$=0.5 \,m\,m$

Hence, the electron has a helical trajectory of radius $0.5 \,m\,m$, with axis of the solenoid along the magnetic field direction.