Question

In Thomson mass spectrograph $\vec{E}\perp \vec{B} $ then the velocity of electron beam will be

• A. $\frac {|\overline{E}|}{|\overline{B}|}$
• B. $\overline{E}\times \overline{B}$
• C. $\frac {|\overline{B}|}{|\overline{E}|}$
• D. $\frac {\overline{E^2}}{\overline{B^2}}$

Answer 1

The Correct Option is A

To determine the velocity of an electron beam in a Thomson mass spectrograph when the electric field $\vec{E}$ is perpendicular to the magnetic field $\vec{B}$, we can use basic principles of electromagnetism.

In this scenario, the electric field and magnetic field are applied perpendicular to each other and also to the path of the electron beam. In such cases, the electric force and magnetic force on the electrons should be equal and opposite for the electrons to pass through undeflected.

1. The electric force acting on an electron is given by: $F_E = e|\overline{E}|$, where $e$ is the charge of the electron.
2. The magnetic force acting on an electron moving with velocity $v$ is: $F_B = e|\overline{B}|v$.
3. For the electron beam to pass through undeflected, these forces should be equal: $F_E = F_B$.
4. Equating the forces, we get: $e|\overline{E}| = e|\overline{B}|v$.
5. We can cancel $e$ from both sides to find the velocity: $|\overline{E}| = |\overline{B}|v$.
6. Solving for $v$, we get: $v = \frac{|\overline{E}|}{|\overline{B}|}$.

Therefore, the correct answer is $\frac{|\overline{E}|}{|\overline{B}|}$, which is option 1.

Let's rule out other options:

• $\overline{E} \times \overline{B}$: This represents a vector cross product which would give a vector quantity, not a velocity.
• $\frac{|\overline{B}|}{|\overline{E}|}$: This would imply the opposite ratio for the velocity.
• $\frac{\overline{E^2}}{\overline{B^2}}$: This represents a square of fields divided by square of fields, not pertaining to velocity calculation.