Question

A narrow electron beam passes undeviated through an electric field of \( 4\times10^{4}\ \text{V/m} \) and a magnetic field \( 2\times10^{-3}\ \text{weber/m}^2 \) applied at the same place. The speed of the electron beam will be:

• A. \( 2\times10^{7}\ \text{m/s} \)
• B. \( 8\times10^{4}\ \text{m/s} \)
• C. \( 6\times10^{6}\ \text{m/s} \)
• D. \( 6\times10^{7}\ \text{m/s} \)

Hint:

For an undeviated beam in crossed fields the electric and magnetic forces cancel, so \( qE = qvB \Rightarrow v = E/B \).

Answer 1

The Correct Option is A

Step 1: Condition for undeviated motion (velocity selector).
When crossed electric and magnetic fields act on a moving charge and the beam goes straight through, the electric force and the magnetic force balance each other:
\[ qE = qvB. \]
Step 2: Solve for the speed.
The charge \( q \) cancels, giving
\[ v = \frac{E}{B}. \]
Step 3: Substitute the values.
\[ v = \frac{4\times10^{4}\ \text{V/m}}{2\times10^{-3}\ \text{Wb/m}^2}. \]
Step 4: Arithmetic.
\[ v = \frac{4}{2}\times10^{4-(-3)} = 2\times10^{7}\ \text{m/s}. \]
This matches option (i). The other options do not satisfy \( v = E/B \).

\[\boxed{v = 2\times10^{7}\ \text{m/s}}\]