A charge particle is projected in a region of magnetic field and electric field if mass if mass of charged particle is m a its speed charges from `V_(0)` to `2V_(0) [W_(E)=` work done by electric field `W_(B)=` work done by magnetic field] which statement is incorrect.?

Options

A. `W_(B)=(1)/(2)mv_(0)^(2)`
B. `W_(B)+W_(E)=(3)/(2)mv_(0)^(2)`
C. `W_(E)=(3)/(2)mv_(0)^(2)`
D. `W_(E)-W_(B)=(3)/(2)mv_(0)^(2)`

Explanation

To solve the problem, we need to analyze the situation involving a charged particle moving in both electric and magnetic fields. We will apply the work-energy theorem and consider the properties of the forces involved.

### Step-by-Step Solution:

1. **Identify the Initial and Final Velocities:**
- The initial velocity \( V_0 \) and the final velocity \( 2V_0 \) are given.

2. **Calculate the Change in Kinetic Energy:**
- The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- Initial kinetic energy \( KE_i \):
\[ KE_i = \frac{1}{2} m (V_0)^2 = \frac{1}{2} m V_0^2 \]
- Final kinetic energy \( KE_f \):
\[ KE_f = \frac{1}{2} m (2V_0)^2 = \frac{1}{2} m (4V_0^2) = 2 m V_0^2 \]
- Change in kinetic energy \( \Delta KE \):
\[ \Delta KE = KE_f - KE_i = 2m V_0^2 - \frac{1}{2} m V_0^2 = \frac{4m V_0^2}{2} - \frac{1}{2} m V_0^2 = \frac{3}{2} m V_0^2 \]

3. **Apply the Work-Energy Theorem:**
- The work-energy theorem states:
\[ W_E + W_B = \Delta KE \]
- Here, \( W_E \) is the work done by the electric field and \( W_B \) is the work done by the magnetic field.

4. **Analyze the Work Done by the Magnetic Field:**
- The magnetic force on a charged particle is given by:
\[ F_B = q(\mathbf{v} \times \mathbf{B}) \]
- The magnetic force is always perpendicular to the velocity of the particle. Therefore, the work done by the magnetic field \( W_B \) is:
\[ W_B = 0 \]
- This is because work is defined as \( W = F \cdot d \), and since the force is perpendicular to the displacement, the work done is zero.

5. **Substituting into the Work-Energy Equation:**
- Since \( W_B = 0 \), we have:
\[ W_E + 0 = \Delta KE \implies W_E = \Delta KE = \frac{3}{2} m V_0^2 \]

6. **Identify the Incorrect Statement:**
- Since \( W_B = 0 \) and \( W_E = \frac{3}{2} m V_0^2 \), any statement claiming that the work done by the magnetic field is non-zero or that the work done by the electric field is not equal to the change in kinetic energy would be incorrect.

### Conclusion:
The incorrect statement is that the work done by the magnetic field \( W_B \) is non-zero.