An electron is moving with speed `2xx10^5 m//s` along the positive x-direction in the presence of magnetic induction `vecB = (hat i + 4 hat j - 3 hat k) T`. The magnitude of the force experienced by the electron in `N ( e = 1.6xx 10^(-19) C) ( vec F = q ( vec v xx vec B))`

Options

A. `1.18xx10^(-13)`
B. `1.28xx10^(-13)`
C. `1.6xx10^(-13)`
D. `1.72xx10^(-13)`

Answer

To solve the problem step by step, we will follow the formula for the magnetic force experienced by a charged particle moving in a magnetic field, which is given by:

\[ \vec{F} = q (\vec{v} \times \vec{B}) \]

Where:
- \( \vec{F} \) is the magnetic force,
- \( q \) is the charge of the particle,
- \( \vec{v} \) is the velocity vector of the particle,
- \( \vec{B} \) is the magnetic field vector.

### Step 1: Identify the given values

- Charge of the electron, \( q = -1.6 \times 10^{-19} \, \text{C} \) (we will use the magnitude for calculations).
- Velocity of the electron, \( \vec{v} = 2 \times 10^5 \, \hat{i} \, \text{m/s} \).
- Magnetic field, \( \vec{B} = \hat{i} + 4\hat{j} - 3\hat{k} \, \text{T} \).

### Step 2: Calculate the cross product \( \vec{v} \times \vec{B} \)

To find \( \vec{F} \), we first need to calculate \( \vec{v} \times \vec{B} \).

\[ \vec{v} \times \vec{B} = (2 \times 10^5 \hat{i}) \times (\hat{i} + 4\hat{j} - 3\hat{k}) \]

Using the distributive property of the cross product:

\[ = (2 \times 10^5 \hat{i}) \times \hat{i} + (2 \times 10^5 \hat{i}) \times (4\hat{j}) + (2 \times 10^5 \hat{i}) \times (-3\hat{k}) \]

Calculating each term:

1. \( \hat{i} \times \hat{i} = 0 \)
2. \( \hat{i} \times \hat{j} = \hat{k} \) → \( 2 \times 10^5 \times 4 \hat{k} = 8 \times 10^5 \hat{k} \)
3. \( \hat{i} \times \hat{k} = -\hat{j} \) → \( 2 \times 10^5 \times (-3) \hat{j} = -6 \times 10^5 \hat{j} \)

Combining these results:

\[ \vec{v} \times \vec{B} = 0 + 8 \times 10^5 \hat{k} - 6 \times 10^5 \hat{j} = -6 \times 10^5 \hat{j} + 8 \times 10^5 \hat{k} \]

### Step 3: Calculate the force \( \vec{F} \)

Now we can calculate \( \vec{F} \):

\[ \vec{F} = q (\vec{v} \times \vec{B}) = -1.6 \times 10^{-19} \times (-6 \times 10^5 \hat{j} + 8 \times 10^5 \hat{k}) \]

Distributing \( q \):

\[ \vec{F} = 1.6 \times 10^{-19} \times 6 \times 10^5 \hat{j} - 1.6 \times 10^{-19} \times 8 \times 10^5 \hat{k} \]

Calculating each component:

1. \( 1.6 \times 6 = 9.6 \)
2. \( 1.6 \times 8 = 12.8 \)

Thus, we have:

\[ \vec{F} = 9.6 \times 10^{-14} \hat{j} - 12.8 \times 10^{-14} \hat{k} \]

### Step 4: Calculate the magnitude of the force

The magnitude of the force vector \( \vec{F} \) is given by:

\[ |\vec{F}| = \sqrt{(9.6 \times 10^{-14})^2 + (-12.8 \times 10^{-14})^2} \]

Calculating:

\[ = \sqrt{(9.6^2 + 12.8^2) \times 10^{-28}} \]

Calculating \( 9.6^2 + 12.8^2 \):

\[ 9.6^2 = 92.16, \quad 12.8^2 = 163.84 \]

Adding these:

\[ 92.16 + 163.84 = 256 \]

Thus:

\[ |\vec{F}| = \sqrt{256 \times 10^{-28}} = 16 \times 10^{-14} = 1.6 \times 10^{-13} \, \text{N} \]

### Final Answer

The magnitude of the force experienced by the electron is:

\[ |\vec{F}| = 1.6 \times 10^{-13} \, \text{N} \]