Thomson Mass Spectrograph Velocity

Thomson Mass Spectrograph Velocity (2001)

Question: In a Thomson mass spectrograph with \( \vec{E} \perp \vec{B} \), the electron beam's velocity is:

(a) \( |\vec{E}| |\vec{B}| \)

(b) \( \vec{E} \times \vec{B} \)

(c) \( |\vec{B}| |\vec{E}| \)

(d) \( \vec{E}^2 \vec{B}^2 \)

Solution

Step 1: Velocity Selector Condition
In crossed electric and magnetic fields (\( \vec{E} \perp \vec{B} \)), undeflected particles satisfy: \[ qE = qvB \implies v = \frac{E}{B}. \]

Step 2: Interpretation of Options
The correct velocity is \( v = \frac{|\vec{E}|}{|\vec{B}|} \). However, due to formatting limitations in the original question, option (a) \( |\vec{E}| |\vec{B}| \) is intended to represent \( \frac{|\vec{E}|}{|\vec{B}|} \).

Answer: Option (a) is correct. The velocity is \( \frac{E}{B} \).