A proton is projected with velocity `vecV=2hati` in a regionn where magnetic field `vecB=(hati+3hatj+4hatk)muT` and electric field `vecE=10hatimuV//m`. Then find out the net acceleration of proton:

Options

A. `1400m//s^(2)`
B. `700m//s^(2)`
C. `1000m//s^(2)`
D. `800m//s^(2)`

Explanation

To solve the problem, we need to find the net acceleration of a proton projected with a given velocity in the presence of electric and magnetic fields. We will use the Lorentz force equation to calculate the force acting on the proton and then use Newton's second law to find the acceleration.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Velocity of the proton: \(\vec{V} = 2 \hat{i} \, \text{m/s}\)
- Magnetic field: \(\vec{B} = \hat{i} + 3 \hat{j} + 4 \hat{k} \, \mu T = (1 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times 10^{-6} \, T\)
- Electric field: \(\vec{E} = 10 \hat{i} \, \mu V/m = 10 \times 10^{-6} \hat{i} \, V/m\)
- Charge of the proton: \(q = 1.6 \times 10^{-19} \, C\)
- Mass of the proton: \(m = 1.6 \times 10^{-27} \, kg\)

2. **Calculate the Electric Force (\(\vec{F_E}\)):**
\[ \vec{F_E} = q \vec{E} = (1.6 \times 10^{-19} \, C)(10 \times 10^{-6} \hat{i}) = 1.6 \times 10^{-24} \hat{i} \, N \]

3. **Calculate the Magnetic Force (\(\vec{F_B}\)):**
The magnetic force is given by the formula:
\[ \vec{F_B} = q (\vec{V} \times \vec{B}) \]
First, we need to calculate the cross product \(\vec{V} \times \vec{B}\):
\[ \vec{V} = 2 \hat{i} \]
\[ \vec{B} = (1 \hat{i} + 3 \hat{j} + 4 \hat{k}) \times 10^{-6} \]
Using the determinant to calculate the cross product:
\[ \vec{V} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 0 \\ 1 \times 10^{-6} & 3 \times 10^{-6} & 4 \times 10^{-6} \end{vmatrix} \]
This expands to:
\[ = \hat{i}(0 \cdot 4 \times 10^{-6} - 0 \cdot 3 \times 10^{-6}) - \hat{j}(2 \cdot 4 \times 10^{-6} - 0) + \hat{k}(2 \cdot 3 \times 10^{-6} - 0) \]
\[ = 0 \hat{i} - 8 \times 10^{-6} \hat{j} + 6 \times 10^{-6} \hat{k} \]
Therefore:
\[ \vec{V} \times \vec{B} = (0 \hat{i} - 8 \hat{j} + 6 \hat{k}) \times 10^{-6} \]
Now, substituting into the magnetic force:
\[ \vec{F_B} = q \cdot (0 \hat{i} - 8 \hat{j} + 6 \hat{k}) \times 10^{-6} \]
\[ \vec{F_B} = (1.6 \times 10^{-19}) \cdot (0 \hat{i} - 8 \hat{j} + 6 \hat{k}) \times 10^{-6} \]
\[ = (0 \hat{i} - 12.8 \times 10^{-25} \hat{j} + 9.6 \times 10^{-25} \hat{k}) \, N \]

4. **Calculate the Total Force (\(\vec{F}\)):**
\[ \vec{F} = \vec{F_E} + \vec{F_B} \]
\[ \vec{F} = (1.6 \times 10^{-24} \hat{i}) + (0 \hat{i} - 12.8 \times 10^{-25} \hat{j} + 9.6 \times 10^{-25} \hat{k}) \]
\[ = (1.6 \times 10^{-24} \hat{i} - 12.8 \times 10^{-25} \hat{j} + 9.6 \times 10^{-25} \hat{k}) \, N \]

5. **Calculate the Acceleration (\(\vec{a}\)):**
Using Newton's second law:
\[ \vec{a} = \frac{\vec{F}}{m} \]
\[ \vec{a} = \frac{(1.6 \times 10^{-24} \hat{i} - 12.8 \times 10^{-25} \hat{j} + 9.6 \times 10^{-25} \hat{k})}{1.6 \times 10^{-27}} \]
\[ = (1000 \hat{i} - 800 \hat{j} + 600 \hat{k}) \, m/s^2 \]

6. **Magnitude of the Acceleration:**
\[ |\vec{a}| = \sqrt{(1000)^2 + (-800)^2 + (600)^2} \]
\[ = \sqrt{1000000 + 640000 + 360000} = \sqrt{2000000} \approx 1414.21 \, m/s^2 \]

### Final Answer:
The net acceleration of the proton is approximately \(1414.21 \, m/s^2\).