An electron beam passes through a magnetic field of `2xx10^(-3) Wb//m^(2)` and an electric field of `3.2xx10^(4) V//m`, both acting simultaneously.`(vecE_|_vecB_|_vecV)` If the path of electrons remains undeflected calculate the speed of [mass of electron`=9.1xx10^(-31)kg`]?

Explanation

To solve the problem step by step, we need to understand the conditions under which the electron beam remains undeflected while passing through the electric and magnetic fields.

### Step 1: Understand the Forces Acting on the Electron
When an electron moves through an electric field (E) and a magnetic field (B) simultaneously, it experiences two forces:
1. Electric Force (F_E) = qE
2. Magnetic Force (F_B) = q(v × B)

For the electron beam to remain undeflected, these two forces must be equal in magnitude and opposite in direction:
\[ F_E = F_B \]

### Step 2: Set Up the Equation
Since the forces are equal, we can write:
\[ qE = qvB \]
Here, q is the charge of the electron, E is the electric field strength, v is the speed of the electron, and B is the magnetic field strength.

### Step 3: Cancel the Charge
Since the charge (q) appears on both sides of the equation, we can cancel it out (assuming it is non-zero):
\[ E = vB \]

### Step 4: Solve for the Speed of the Electron (v)
Rearranging the equation gives us:
\[ v = \frac{E}{B} \]

### Step 5: Substitute the Given Values
Now we substitute the given values for the electric field (E) and the magnetic field (B):
- E = \( 3.2 \times 10^4 \, \text{V/m} \)
- B = \( 2 \times 10^{-3} \, \text{Wb/m}^2 \)

Substituting these values into the equation:
\[ v = \frac{3.2 \times 10^4}{2 \times 10^{-3}} \]

### Step 6: Calculate the Speed
Now, perform the calculation:
\[ v = \frac{3.2 \times 10^4}{2 \times 10^{-3}} = \frac{3.2}{2} \times 10^{4 + 3} = 1.6 \times 10^6 \, \text{m/s} \]

### Final Answer
Thus, the speed of the electron is:
\[ v = 1.6 \times 10^6 \, \text{m/s} \]

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