1. Given data

• Initial kinetic energy of electron: $K = 100\text{ eV} = 100(1.6\times 10^{-19})\text{ J} = 1.6\times 10^{-17}\text{ J}$
• Magnetic field: $\vec B = 1.5\times 10^{-3}\,\hat k\,\text{T}$
• Electron enters along the $x$-axis.
• Magnetic field region: from $x=0$ to $x=2\text{ cm}$.
• Screen is $8\text{ cm}$ away from $S$.

So after leaving the field region at $x=2\text{ cm}$, the electron travels another $8-2=6\text{ cm}$ without field before hitting the screen.

2. Speed of the electron

Using $K=\frac12 mv^2$ we get $v=\sqrt{\frac{2K}{m}}$

Substitute values: $v=\sqrt{\frac{2(1.6\times 10^{-17})}{9.1\times 10^{-31}}}$ $v=\sqrt{3.516\times 10^{13}}\approx 5.93\times 10^6\,\text{m/s}$

3. Radius of circular path inside magnetic field

For motion perpendicular to $\vec B$, $r=\frac{mv}{eB}$

Substitute: $r=\frac{9.1\times 10^{-31}\times 5.93\times 10^6}{1.6\times 10^{-19}\times 1.5\times 10^{-3}}$

$r\approx 2.25\times 10^{-2}\text{ m}=2.25\text{ cm}$

4. Angle turned inside the magnetic field region

The electron enters at $x=0$ and leaves the field at $x=2\text{ cm}$.

Inside the field, it moves on a circular arc of radius $r=2.25\text{ cm}$. If the angular deflection is $\theta$, then horizontal advance in the field is $r\sin\theta = 2\text{ cm}$

So, $\sin\theta = \frac{2}{2.25} = 0.8889$ $\theta \approx 62.7^\circ$

Also, $\cos\theta = \sqrt{1-\sin^2\theta} \approx \sqrt{1-0.8889^2} \approx 0.4581$

5. Vertical displacement while inside the field

The electron bends downward/upward depending on sign, but we only need magnitude. The displacement from $S$ to exit point is $y_1 = r(1-\cos\theta)$

Thus, $y_1 = 2.25(1-0.4581)\text{ cm}$ $y_1 \approx 2.25(0.5419)=1.22\text{ cm}$

6. Motion after leaving the field

After exiting, the electron moves in a straight line tangent to the circular path. The tangent makes angle $\theta$ with the $x$-axis.

Remaining horizontal distance to screen: $6\text{ cm}$

So additional vertical displacement is $y_2 = 6\tan\theta$

Now, $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{0.8889}{0.4581} \approx 1.940$

Hence, $y_2 = 6\times 1.940 = 11.64\text{ cm}$

7. Total displacement on the screen

Therefore, $d = y_1 + y_2 = 1.22 + 11.64 = 12.86\text{ cm}$

So, $\boxed{d \approx 12.87\text{ cm}}$

This matches Option B.

8. Comparison with stored answer

Stored correct answer: B

Derived answer: B

They agree.