A charged particle with a velocity `2xx10^(3) ms^(-1)` passes undeflected through electric field and magnetic fields in mutually perpendicular directions. The magnetic field is `1.5 T. the magnitude of electric field will be

Options

A. `1.5xx10^(3) NC^(-1)`
B. `2xx10^(3)NC^(-1)`
C. `3xx10^(3)NC^(-1)`
D. `1.33xx10^(3)NC^(-1)`

Answer

To solve the problem, we need to find the magnitude of the electric field (E) when a charged particle moves through mutually perpendicular electric (E) and magnetic (B) fields without being deflected.

### Step-by-Step Solution:

1. **Understanding the Forces**:
- When a charged particle moves through an electric field, it experiences an electric force (F_E) given by:
\[ F_E = qE \]
- When it moves through a magnetic field, it experiences a magnetic force (F_M) given by:
\[ F_M = q(v \times B) \]
- Since the electric and magnetic fields are mutually perpendicular, we can simplify the magnetic force to:
\[ F_M = qvB \]

2. **Condition for Undeflected Motion**:
- For the particle to pass undeflected, the magnitudes of the electric force and magnetic force must be equal:
\[ F_E = F_M \]
- Therefore, we can equate the two forces:
\[ qE = qvB \]

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

4. **Substituting Known Values**:
- We are given:
- Velocity, \( v = 2 \times 10^3 \, \text{m/s} \)
- Magnetic field, \( B = 1.5 \, \text{T} \)
- Now substituting these values into the equation:
\[ E = (2 \times 10^3 \, \text{m/s}) \times (1.5 \, \text{T}) \]

5. **Calculating the Electric Field**:
- Performing the multiplication:
\[ E = 3 \times 10^3 \, \text{N/C} \]

### Final Answer:
The magnitude of the electric field is \( E = 3 \times 10^3 \, \text{N/C} \).

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