An electron is projected horizontally with a kinetic energy of 10...

Kinetic energy of the electron, $K = 10 \text{ keV} = 10 \times 10^3 \times 1.6 \times 10^{-19} \text{ J} = 1.6 \times 10^{-15} \text{ J}$

The radius of the circular path described by the electron in the magnetic field is given by:

$R = \dfrac{mv}{qB} = \dfrac{\sqrt{2mK}}{qB}$

Substituting the given values:

$R = \dfrac{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-15}}}{1.6 \times 10^{-19} \times 1.0 \times 10^{-7}}$

$R = \dfrac{\sqrt{29.12 \times 10^{-46}}}{1.6 \times 10^{-26}} = \dfrac{5.396 \times 10^{-23}}{1.6 \times 10^{-26}} \approx 3372.5 \text{ m}$

Since the distance travelled $x = 1 \text{ m}$ is much smaller than the radius $R$, the lateral deflection $y$ can be approximated using the relation $y \approx \dfrac{x^2}{2R}$:

$y \approx \dfrac{1^2}{2 \times 3372.5} \approx 1.48 \times 10^{-4} \text{ m}$

$y \approx 0.0148 \text{ cm} \approx 0.015 \text{ cm}$