A beam of charged particle enters in a region of magnetic field of `5 xx 10^(-3)` weber m and electric field of `2.5 xx 10^(4) Vm^(1)`,acting perpendicularly. Calculate the speed of particles perpendicular to electric and ,magnetic field , if their path remains unchanged. The given charge on particles and mass are ` 3.2 xx 10^(-19) C " and " 12 xx 10^(-31) ` kg , respectively.

Answer

To solve the problem, we need to find the speed of charged particles that enter a region with both an electric field and a magnetic field acting perpendicularly to each other, such that the path of the particles remains unchanged.

### Given Data:
- Magnetic field, \( B = 5 \times 10^{-3} \, \text{Wb/m}^2 \)
- Electric field, \( E = 2.5 \times 10^{4} \, \text{V/m} \)
- Charge of the particle, \( q = 3.2 \times 10^{-19} \, \text{C} \)
- Mass of the particle, \( m = 12 \times 10^{-31} \, \text{kg} \)

### Step-by-Step Solution:

1. **Understand the Forces Acting on the Particle**:
When a charged particle moves in a magnetic field, it experiences a magnetic force given by:
\[ F_B = qvB \]
where \( v \) is the velocity of the particle.

In the presence of an electric field, the particle also experiences an electric force given by:
\[ F_E = qE \]

2. **Condition for Unchanged Path**:
For the path of the charged particle to remain unchanged, the magnetic force must be equal to the electric force:
\[ F_B = F_E \]
Therefore, we can set the two forces equal to each other:
\[ qvB = qE \]

3. **Cancel the Charge**:
Since \( q \) is non-zero, we can divide both sides of the equation by \( q \):
\[ vB = E \]

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

5. **Substitute the Given Values**:
Now, we can substitute the values of \( E \) and \( B \):
\[ v = \frac{2.5 \times 10^{4} \, \text{V/m}}{5 \times 10^{-3} \, \text{Wb/m}^2} \]

6. **Calculate the Velocity**:
Performing the division:
\[ v = \frac{2.5 \times 10^{4}}{5 \times 10^{-3}} = 0.5 \times 10^{7} \, \text{m/s} \]

7. **Final Result**:
Thus, the speed of the particles is:
\[ v = 5 \times 10^{6} \, \text{m/s} \]

### Summary of the Solution:
The speed of the charged particles, when their path remains unchanged in the presence of perpendicular electric and magnetic fields, is \( 5 \times 10^{6} \, \text{m/s} \).