In a certain region of space electric field E and magnetic field B are perpendicular to each other and an electron enters region perpendicular to the direction of B and E both and moves undeflected, then velocity of electron is
(a) `(|E|)/(|B|)` (b) `ExxB` (c) `(|B|)/(|E|)` (d) `E.B`

Options

A. `(|E|)/(|B|)`
B. `ExxB`
C. `(|B|)/(|E|)`
D. `E.B`

Answer

To solve the problem, we need to analyze the forces acting on the electron when it enters a region with both electric and magnetic fields. The key points to consider are:

1. **Understanding the Forces**: The electron experiences two forces:
- The electric force (\( F_E \)) due to the electric field (\( E \)):
\[ F_E = qE \]
- The magnetic force (\( F_B \)) due to the magnetic field (\( B \)):
\[ F_B = q(v \times B) \]
Here, \( q \) is the charge of the electron, \( v \) is its velocity, and \( \times \) denotes the cross product.

2. **Condition for Undeflected Motion**: For the electron to move undeflected, these two forces must be equal in magnitude and opposite in direction:
\[ F_E = F_B \]
This leads to the equation:
\[ qE = q(v \times B) \]

3. **Canceling the Charge**: Since the charge \( q \) appears on both sides of the equation, we can cancel it out (assuming \( q \neq 0 \)):
\[ E = vB \]

4. **Solving for Velocity**: Rearranging the equation gives us the expression for the velocity \( v \):
\[ v = \frac{E}{B} \]

5. **Identifying the Correct Option**: The expression we derived matches option (a):
\[ v = \frac{|E|}{|B|} \]

Thus, the velocity of the electron is given by:
\[ \text{Answer: } v = \frac{|E|}{|B|} \]