GSO003 Problem 7

1. Velocity Selector (Step 1)

The electron velocity is $\vec{v} = (v_0, 0, 0)$ .

Electric force ( $-y$ direction): $\vec{F}_E = -e\vec{E} = (0, -eE_0, 0)$

Magnetic (Lorentz) force ( $+y$ direction): $\vec{F}_B = -e(\vec{v} \times \vec{B}) = (0, ev_0B_0, 0)$

For straight-line motion, these must balance:

-eE_0 + ev_0B_0 = 0 \implies v_0 = \frac{E_0}{B_0}

2. Exact Circular Orbit in the Magnetic Field (Step 2)

With the electric field off, the electron undergoes uniform circular motion. The radius $R$ satisfies:

m\frac{v_0^2}{R} = ev_0B_0 \implies R = \frac{mv_0}{eB_0}

The electron starts at the origin moving in the $+x$ direction and curves toward $+y$ , so the center of curvature is at $(0, R)$ . The circular orbit equation is:

x^2 + (y - R)^2 = R^2

At $x = L$ , the exit $y$ -coordinate $y_1$ (with $y_1 < R$ ) is:

y_1 = R - \sqrt{R^2 - L^2} = R(1 - \cos\theta)

where $\theta$ is the deflection angle satisfying:

\sin\theta = \frac{L}{R}, \quad \cos\theta = \frac{\sqrt{R^2 - L^2}}{R}

Using the half-angle formula ( $1 - \cos\theta = 2\sin^2(\theta/2)$ , $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$ ):

y_1 = \frac{L(1 - \cos\theta)}{\sin\theta} = L\tan\frac{\theta}{2}

3. Drift Space and Total Displacement (Step 3)

In the field-free region, the electron travels at angle $\theta$ to the $x$ -axis. Additional $y$ -displacement over distance $D$ :

y_2 = D\tan\theta

Total displacement on the screen:

Y = y_1 + y_2 = L\tan\frac{\theta}{2} + D\tan\theta

4. Determining the Deflection Angle

Substituting $D = \sqrt{3}L$ and $Y = (3 - \sqrt{3})L$ :

\tan\frac{\theta}{2} + \sqrt{3}\tan\theta = 3 - \sqrt{3}

The left side is strictly increasing for $0 < \theta < \pi/2$ , so the solution is unique.

Testing $\theta = 30^\circ$ (using $\tan 30^\circ = 1/\sqrt{3}$ , $\tan 15^\circ = 2 - \sqrt{3}$ ):

(2 - \sqrt{3}) + \sqrt{3} \cdot \frac{1}{\sqrt{3}} = 2 - \sqrt{3} + 1 = 3 - \sqrt{3} \checkmark

Therefore $\theta = 30^\circ$ is the unique solution.

5. Computing the Specific Charge

From $\sin 30^\circ = 1/2$ :

\frac{L}{R} = \frac{1}{2} \implies R = 2L

Using $R = \frac{mv_0}{eB_0}$ and $v_0 = \frac{E_0}{B_0}$ :

R = \frac{mE_0}{eB_0^2} \implies \frac{e}{m} = \frac{E_0}{RB_0^2} = \frac{E_0}{2LB_0^2}

Substituting numerical values:

2LB_0^2 = 2 \times (5.0 \times 10^{-2}) \times (1.0 \times 10^{-3})^2 = 10^{-7}

\frac{e}{m} = \frac{1.76 \times 10^4}{10^{-7}} = 1.76 \times 10^{11}\,\mathrm{C/kg}

$A = 1.76$ , so $100A = 176$ .

Answer: 176

GSO003 Problem 7

Thomson Experiment with Exact Trajectory Geometry

Problem Statement

Constraints

Input Format

Solution

1. Velocity Selector (Step 1)

2. Exact Circular Orbit in the Magnetic Field (Step 2)

3. Drift Space and Total Displacement (Step 3)

4. Determining the Deflection Angle

5. Computing the Specific Charge