If the acceleration and velocity of a charged particle moving in a constnt magnetic region is given by `a=a_1hati+a_2hatk, v=b_1hati+b_2hatk.[a_1,a_2,b_1` and `b_2` are constant]. Then choose the wrong statement

Options

A. magnetic field may be along y-axis
B. `a_1b_1+a_2b_2=0`
C. magnetic field is along x-axis
D. kinetic energy of particle is always constant

Explanation

To solve the problem, we need to analyze the statements regarding the motion of a charged particle in a magnetic field, given its acceleration and velocity vectors.

### Step-by-Step Solution:

1. **Understanding the Given Vectors**:
- The acceleration vector \( \vec{a} \) is given as:
\[ \vec{a} = a_1 \hat{i} + a_2 \hat{k} \]
- The velocity vector \( \vec{v} \) is given as:
\[ \vec{v} = b_1 \hat{i} + b_2 \hat{k} \]

2. **Magnetic Force and Motion**:
- The force experienced by a charged particle moving in a magnetic field is given by:
\[ \vec{F} = Q (\vec{B} \times \vec{v}) \]
- According to Newton's second law, this force can also be expressed as:
\[ \vec{F} = m \vec{a} \]
- This leads to the equation:
\[ m \vec{a} = Q (\vec{B} \times \vec{v}) \]

3. **Perpendicularity of Vectors**:
- The cross product \( \vec{B} \times \vec{v} \) results in a vector that is perpendicular to both \( \vec{B} \) and \( \vec{v} \). Therefore, \( \vec{a} \) is also perpendicular to \( \vec{v} \).
- This means:
\[ \vec{a} \cdot \vec{v} = 0 \]

4. **Calculating the Dot Product**:
- Substituting the expressions for \( \vec{a} \) and \( \vec{v} \):
\[ (a_1 \hat{i} + a_2 \hat{k}) \cdot (b_1 \hat{i} + b_2 \hat{k}) = a_1 b_1 + a_2 b_2 = 0 \]
- This shows that the statement \( a_1 b_1 + a_2 b_2 = 0 \) is true.

5. **Analyzing the Statements**:
- **Statement A**: "Magnetic field may be along y-axis." - This can be true since \( \vec{B} \) can be perpendicular to both \( \vec{a} \) and \( \vec{v} \).
- **Statement B**: "a1b1 + a2b2 = 0." - This is true as derived above.
- **Statement C**: "Magnetic field is along x-axis." - This is likely false because for \( \vec{B} \) to be along the x-axis, both \( \vec{a} \) and \( \vec{v} \) must have components in the y and z directions, which contradicts the perpendicularity condition.
- **Statement D**: "Kinetic energy of the particle is always constant." - This is true because magnetic forces do not do work on the charged particle.

6. **Conclusion**:
- The wrong statement is **Statement C**: "Magnetic field is along x-axis."