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: Recognise the setup as a velocity selector.
Perpendicular \(E\) and \(B\) fields let only one particular speed pass in a straight line. For that speed the two sideways forces are equal and opposite, so the net transverse force is zero.

Step 2: Equate the magnitudes of the forces.
Electric force magnitude \(= eE\); magnetic force magnitude \(= evB\). Setting them equal: \(eE = evB\), and cancelling the electron charge \(e\) leaves \(E = vB\).

Step 3: Rearrange and insert numbers.
\(v = \dfrac{E}{B} = \dfrac{4\times10^{4}}{2\times10^{-3}}\). Handle the mantissa and powers of ten separately: \(4/2 = 2\) and \(10^{4}/10^{-3} = 10^{7}\).

Step 4: Combine.
\(v = 2\times10^{7}\) m/s, so option (i) is correct.

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